Learn how to find the unknown angle x in this triangle. Use the Exterior Angle Theorem and the Straight Angle Property. Step-by-step tutorial by PreMath.com
Sir, beautifully & patiently explained. You have an ocean of patience which is the prime requirement of any teacher Even a student weak in Maths will easily understand if u take a class Thanks
Sir, you are truly the Sherlock Holmes of trigonometry. You assemble the evidence point by point, and arrive at the solution step by step. It is making mathematics good fun! Thank you.
Nice geometric solution! I got there with tangents function. If you complete the right-angled triangle in the bottom left, with height h and base p+2d (were d is the length AD) then you see there are three right-angled triangles and by considering each one in turn that: tan 60 = (p+2d)/h = sqrt(3) tan 45 = (p+d)/h = 1 tan (45-x) = p/h Multiply both sides of the second equation by two and subtract the first equation from it and you have 2-sqrt(3) = p/h Substitute this into the third equation and you get: x = 45 - tan-1 (2-sqrt(3)) which is 30 degrees.
Let's say that you are required to solve by construction only, no trigonometry. Eastern Brown's solution adapts as follows. Complete the right triangle and make use of the known ratios of sides for the 60°-30°-90°, 45°-45°-90°, and 75°-15°-90° right triangles. The smallest right triangle has p/h = (2-√3) = (√3-1)/(√3+1) which matches the 75°-15°-90° right triangle. The smallest angle, 15°, needs to be subtracted from 45° to get x = 30°.
In 1970, my Maths teacher at Whitchurch high school in Cardiff, Mr Smth, was so good he helped me to learn how to do this stuff and I am forever grateful. I still haven't found a use for the integration of 1 + Tan squared (x) though. :)
Great puzzle- I needed the DP portion to start before I could solve it. Thanks for providing such good content and a mix of easier and more difficult problems.
After watching few similar videos able to solve this one in mind. It's mostly in finding/drawing the correct isosceles, equilateral or congruent triangles with the equal length sides.
Another flash of brilliance from the maestro, that was a superb demonstration, I'd never have got that, I like the way in many of your marvels that you add cunning lines that reveal the pathway, it's a joy to watch. Thanks again 👍🏻
Don't know what your age or former occupational title has to do with it. And you "guessed" it correctly? There's no guessing as you couldn't possibly know if you "guess" was correct until someone else DEMOSTRATED it to be correct. Ugh You must have been an excellent engineer. SMH
Thanks for the excellent explanation. I stared at this problem for several minutes before coming up with the answer. It seemed obvious to me that angle X should be equal to angle ABC and ABC was obviously 30. based on the given angles, but while I could prove angle ABC couldn't figure out how to prove angle X.
Before these things used to be a headache for me in class. I did it cause I was forced to and did the minimum. Now it’s my entertainment and I can’t get enough of it. I rather watch this than a series.
Interesting solution. Another way is drawing CH and proving that A and H are one point, because the other two possible positions of H lead to contradictions with the triangle CHB.
Beautiful solution to the problem. Very much in the domain where math is art itself. I have opted for a cruder way (PS: I am an engineer). I dropped a perpendicular instead and used expressions for tan. Got two equations and soleved simultaneously to get the answer.
It's over 30 years since I took geometry in university (a Euclidian and non-Euclidian course). I thought it was not possible to find X. I also misinterpreted the double hashes as meaning AD and DB were parallel :D So I didn't get far. I was amazing by the explanation. Such a smart tactic. Thanks for taking the time.
I used the law of sines and cosines. Since line AD = DB it's easier to make an assumption about the length and then we can workout CD using the sine rule and CA using the cosine rule then we can use the sine rule again to find the value of X. But thank you, this was a good exercise.
Exactly.. and using this method, you can find that 2x = 60 in less than 1.5 minutes. That's way more than the allotted time for solving a geometry problem like this in CAT exam. We usually have 25-30 seconds for solving such problems in CAT exam.
I solved by using trignometry and the law of sines for triangles. With reference to your diagram, let AD=BD= a, and let us designate the length AC=b. Then, for the triangle ACD, a/sinx = b/sin45 ==> sinx = a/(b*sqrt 2). Also for the triangle ACB, 2a/sin(x+15) = b/sin 30 ==> sin(x+15) = a/b = sqrt 2*sinx. Therefore, sinx*cos15+cosx*sin15 = sqrt 2*sinx. If we divide this equation by sinx, it can be reduced to cos15 + cotx*sin15 = sqrt 2 ==> cotx = (sqrt 2 - cos15)/sin15, or tan x = sin15/(sqrt 2 - cos15). The right hand expression can be calculated as 0.57735026919, which happens to be the tan 30 degrees.
True, but the video shows the fundamental or the manual way ehe, even for those who haven't learned trigonometry still can solve the problem... Nice video ☺️
i guess in these kind of problems only the one who created them can solve them, or you can go through a long journey of trying all possible methods and theoremes to finally conclude it
In ∆CDB, angle B + 15° = 45° [exterior angle is the sum of interior opposite angles] Therefore angle B = 30° Let AD = DB = a and AC = b In ∆ABC by law of sine 2a/sin(x+15°) = b/sin30° = b/(1/2) = 2b => sin (x+15°) = a/b --------(1) In ∆ ADC a/sin x = b/sin45° = b/(1/√2) = b√2 => √2 sin x = a/b ---------(2) From (1) and (2) √2 sin x = sin (x+15°) √2 sin x = sin x cos 15° + cos x sin 15° (√2-cos 15°) sin x = cos x sin15° sin x / cos x = sin 15° /(√2-cos 15°) tan x = sin 15° / (√2-cos 15°) tan x = 1/√3 ( pl. refer Note below) = tan 30° Thus the unknown angle x = 30° Note: sin 15° = sin (45°-30°) = sin 45° cos 30° - cos 45° sin30° = [1/√2] [ √3/2 - 1/2] = (√3-1)/2√2 Similarly cos 15 = (√3+1)/2√2 √2-cos 15 = √2 - (√3+1)/2√2 = (3-√3)/2√2 = √3(√3-1)/2√2 = √3 sin 15 Hence sin 15° / (√2-cos 15°) = 1/√3
an easier solution i think would be to just draw a line say CE parallel to AD now using alternate interior angle property we can conclude that angle ADC = angle DCE. Now since the line CE is also parallel to DB therefore we can conclude using alternate angle property that angle DBC = angle BCE. Now angle DCE= angle DCB + angle BCE = 45. Therefore, 15 + BCE = 45, therefore BCE = DBC = 30
I admit I got stuck after step one. designating point P was a clever trick, but only worked out by chance, due to the particular properties of the two triangles. After it turned out that the vertex in D of APD is 60 degrees, the rest was easy!
Intuitively, I figured that the bifurcation of the base created a triangle ADC which you could map to ABC by a reflection across AC, and a rotation around A, followed by a dilation. This would show that angle BAC is the same angle as CAD. The measure of angle BAC I got deriving angle DBC from 180 - 45 = 135, and then 180 - 15 - 135 = 30. As it turns out, this leads to the correct solution, namely x = 30 degrees. But I don't know if this is a fluke :)
3:13 to get angle PDC you don't need the 120 degree. Just use the exterior angle theorem again in the trangle PDC with angle BPD=30degree as the exterior angle, and angle PCD=15degree so angle PDC=30-15=15degree. This is a good question that looks difficult but once the additional lines are drawn it becomes straightforward.
Yes, I found “X”. It was hiding just to the lower right of “C” at the top. It was pretty easy to find. Especially since it was written in red! 😊 Just kidding. I never thought I would use calculus or trigonometry much when I was in high school. Then I got into machining and use it almost daily! Pay attention kids this IS important!
Well, there are many ways of solving this beautiful problem!! 1) using elementary geometry. 2) using sine rule 3) using m-n theorem Using the third method gives u the answer in just 1step...😎😎😉😉 Anyways, thanks for sharing! Love from India!!
M-N theorem is very handy and interesting. Went through its proof. (m+n)cot(theta)=m.cot(alpha)-n.cot(beta). In this problem AD=m DB=n; m=n Theta=135 Alpha = X Beta = 15 Sin(15) = (√3-1)/(2.√2) Cos(15) = (√3+1)/(2.√2) CotX = √3 X = 30. Very nice approach. Today i learnt m-n theorem. Thank you Kusuma I used the second approach
This seems to be one of a number of similar problems that are special situations. We are given a 45 degree ray from the midpoint D of AB and an intersecting ray from A to C. The problem is to find angle ACD. The method of solution only works if the point C is chosen so that angle DCB is 15 degrees or alternatively angle DBC is 30 degrees..
I was just wondering how will you use that information, that those two parts of bottom line is equal, and I'm speechless how beautifully you created a triangle out of it and used that fact... I knew that equal sides of triangle subtend equal opposite angles in a triangle, but i wasn't smart enough to use that fact... Thank you sir, love from India..!
Given CD bisects side AB in equal parts, it would also mean that CD bisects angle C in equal parts. Let me know your thoughts incase you feel I am wrong and why, happy to be corrected.
Very nice approach. I solved it using sine rule. sin(15+x)/2sinx=sin30/sin45=>sin(15+x)/sinx=sin45/sin30. Putting x=30 or 15+x=45 gives the same result x=30. Hence x=30 deg.
*Trigonometric solution:* Drop a perpendicular from C to meet the base line at E. Let CE= a and ∠ECA= θ. Since ∠DCE=45°, we have ED=a, and since ∠EBC= 30°, we have EB=a✓3. This gives AD= (✓3-1)a and EA=ED-AD=(2-√3)a. Now Tanθ=EA/CE = 2-√3, which gives us θ=15°. Hence X=30°.
Clearing away the cobwebs. 55+ years I was solving problems like this as part of a submarine fire-control problem (I.E. what angle to set the torpedo to intersect with the surface target) We had a mechanical analog computer, however, if it failed, you had to know how to do it by hand. Thank you. Narragansett Bay
I got it. My sister told me the triangle CDP is an isosceles because angle PCD and PDC have the same 15 degree. Therefore, PC = PD. The triangle APD is equilateral, hence PC = PA. So angle PCA and PAC are 45 degree each. Consequently, angle ACD is 30 degree. That’s awesome!!! Thank you very much. I had learned geometry 50 years ago. I have to confess this brings back a lot of memories.
Great puzzle! I didn't find this excellent geometry solution. Instead I dropped a perpendicular from C to point E, then used trig to get CE, EA and EB (after setting AD and BD=1). Then I used Pythagoras to get CB=sqrt 2. Then the Law of Sines to get sin x=1/2. Thanks PreMath!
@@hussainfawzer Hello Thanks for the question, I left out a lot of steps and also erred in saying CB=sqrt2. Sorry for that (CA=sqrt2 is correct). To get EA then CE let EA=y and AB=1 (or any number but 1 makes the calculating easier) Since triangle CED is a right triangle with a 45 degree angle, CE=ED=1+y. EB also =2+y Since angle CEB is defined as a right angle and CBE (as shown by PreMath) is 30 degrees, CEB is a 30-60-90 triangle and CB will be twice CE or 2+2y So by Pythagoras CE^2+EB^2=CB^2 or (1+y)^2+(2+y)^2=(2+2y)^2 Simplify and get 2y^2+2y-1=0 and so y=(sqrt3-1)/2 and EC=(sqrt3+1)/2 I hope that answers your question, Hussain. Now one can get CA=sqrt2 from Pythagoras. And looking at triangle CAD, the law of sines gives sinx/1=sin45/sqrt2 and sinx=1/2.
@@waheisel That was beautiful math. It reminded me of a lot of stuff. Ty I used to love math and physics with a dream to be get into cosmology. Due to personal reasons I'm now soon becoming a doctor. It's good but I miss math and physics so badly 😅
@@yuda4626 Thanks for the kind comment Ty. When I was in school I also liked math, physics, and astronomy. I wasn't nearly good enough to make any of those a career. I also became a doctor. Even though you won't use so much math and physics in your career you can still enjoy your daily PreMath puzzle! Best Wishes
THE SHORT: If you could draw DP so that DP=DB, on paper, you could also just as easily draw AP so that AP=AD=CP=PD, on paper... all anchored by the given angles of ADC & DCB. THE LONG: Imagine vertices ABC are 3 stars in a system, with D being a star exactly between star A and star B (let's say with Hubble, we're easily able to measure that the star D is exactly half way, in light years, between stars A & B, all three nicely aligned). From ANY of these given stars, to any other star (in the ABCD set of star system), as well as, angles 15 & 45....nothing was given, could have been given, via assumption or "eye balling".... Everything came about through precise measurements via parallax + trigonometric, standard Candle method, etc. Take one step back and ask yourself: --- what is the astronomical (if we took ABCD to be a set of 4 stars in the Milky Way) or geometric logic that said you could, for example, draw a precisely KNOWN line segment, like DP, from one vertex of BDC, to the opposite side, BC, in such a way that DP is exactly equal to one side, DB.... but NOT be able to do a similar thing, say, AP? After all, AP=AD=DB, DB=DP, and AD=DB.... all anchored by the given angles of 15 (DCB) & 45 (ADC). For example, "P" can NOT be in any other place, along BC, such that AD is not equal to AP (due to the dictates of the angles 15 and 45, as given). In other words, there's neither "wiggle room" for angles ADC & DCB, on the one hand, nor for point "P" along BC such that AP=AD=DB=DP is NOT true, on the other.
During my jee preparation (engineering entrance exam of India) i studied a theorem called m-n cot theorem, and really it gave the answer in just 5-10 seconds :) 😁 BTW great explanation sir😊
@@Z7youtube It's in my comment. Using formula for sine of a difference. sin(135°-x) = sin(135°)cos(x)-cos(135°)sin(x) = (√2/2) cos(x) + (√2/2)sin(x) We used the fact sin(135°)=sin(45°)=√2/2 and cos(135°)=-cos(45°)=-√2/2
could use trig by drawing a perpendicular line from C to P to form right triangle CPA. Triangle CPB is a 30-60-90 right triangle. label the sides as 1, sqrt 3 (or 1.732) and 2 let's label angle CPA as 'y' . y + x =45 degrees and you now have and 45-90-45 with sides 1, 1, and sqrt 2 degree right-triangle line PD =1 and CD= sqrt 2 . Using ASA (15 degree, sqrt 2 and 135 degree) for triangle CDB gives 0.732 for DB , but DB = AD thus PA = 1-0.732 = 0.268. Since the right triangle CPA has the sides 1, 0.268 then using SAS (1, 0.268, and 90 degree) gives 15 degrees for angle y, but y+x=45 degree hence x=30 degree
Is it only me who thought that if AD=DB, CD is the median line of the triangle in C, thus meaning in cuts the angle in half? And if DCB is 15, ACB would be 30?🤔🤔
Actually, that is not always the case, as shown in this problem here. The median of a triangle in a given vertex is not always equal to the bisector of the relative angle. They're the same in some cases, for example if the triangle is equilateral, it is true for alla the vertices.
You're mistaken between median line and angle bisector. This is proven false in this example so if it cit ACB in middle = ACD = DCB =15° which is not the case
Hi, you could have separated triangle cdb and joined AD and DB by rotating clock wise triangle CDB. Making points A and B to be the same point. And you can get the answer in 2 steps. As rotating would create an isosceles triangle. Try it out
Rotating does not create an isosceles triangle. In fact, in the general case, the rotation is more likely to create a 4 sided polygon. In this particular case it creates a triangle, but not an isosceles one. To think this through, think about point C being split by the rotation, where C1 is the original point C, and C2 is created by the rotation. In this case, because angle ADC is 45 degrees and angle CDB is 135 degrees, after the rotation the line segments C1D and DC2 are collinear forming a straight line, with D in the center, since C1D and DC2 are the same length (but since they form a straight line they are not two sides of an isosceles triangle). So one side of the resulting triangle is C1C2, with the other two sides being AC1 and BC2 (A and B are the same point after the rotation). One of the angles in this new triangle is the 15 degrees from angle DCB prior to rotation, another angle is X degrees, formed from angle ACD prior to rotation (now AC1C2 after rotation). We know from the solution presented that X is not 15 degrees but is a value (trying not to spoil) such that none of the angles in the triangle formed by the rotation are the same.
Fantastic and clever solution! I couldn't made it that way! I have another aprroach: 1. Draw the circle that contains point A,B,C. To do so, draw 2 perpendicular lines at the midpoind of AB and BC, these two lines meet at the center O. WE notice that AOB is the angle at the center and the angle ACB (= angle x +15 degrees) is the angle at the circumference. 2. Demonstrate that the angle at the center AOB is equal 90 degrees. Then, we can find the value of the angle ACB = 1/2 AOB= 45 degrees. 3. The angle x= angle ACB - 15 = 45-15= 30 degrees.
@@emaceferli726 Yes, I can. Just label M and N the midpoint of AB and then AC. Draw 2 perpendicular lines to AB, and AC at these points which meet at the center O. Then we have OC=OA=OB, therefore we can draw a circle from the center O with the radius R=OC=OA=OB. It is easy to see that the angle CBA=30 degrees (CMA is the exterior angle=45degrees). The angle CBA is the angle at circumference of which COA is the angle at the center, so COA= 2 CBA=60 degrees----> the triangle COA is an equilateral one. Let's say I is the midpoint of OA. The height from the vertex C to I is also the bisector of the angle ACO (=60 degrees). Because the angle AMC=OMC=45 degrees so CM is the bisector of the angle OMA, so C,I,M are colleniar. Thus the angle x= ACM= 30 degrees
I took Geo Trig last school year and felt very smart knowing how to solve everything. However when I saw this, it was not the case. Thanks for explaining this very thoroughly. It was very fun and interesting to learn how to solve this Geo Challenge. Please make more!
Notice that sin(x)/AD = sin(45)/AC and sin(x+15)/(2*AD) = sin(30)/AC. Combine to get 2*sin(x)/sin(x+15) = sin(45)/sin(30). Upon expansion of sin(x+15) and some algebra, we get x = 30 degrees.
I think the question may be solved by applying basic triangle rules. I tried by following method- 1. By external angle theorem: angle DBC = 30° 2. In triangle DBC, if length of DB is 'a' (against 15°), then CD is 2a (against 30°) 3. Now, in triangle CAD, AD = a, CD = 2a 4. Angle ACD = x, Angle ADC = 45° = y, Angle CAD = z 5. Angle z = 2 (Angle x) [because CD = 2 (AD)]. 6. As, Angle x + y + z = 180° x + 45 + 2(x) = 180° 3x = 180° - 45° = 135° x = 135°/3 x = 45°
There is no rule in a triangle as you wrote "if length of DB is 'a' (against 15°), then CD is 2a (against 30°)". By this info you can just write: CD>AB.
It was an easy puzzle! You could've solved it with a much easy technique. ∠CDA + ∠CDB=180°[Linear Pair] . So, ∠CDA is given as 45°. Then, 45°+ ∠CDB=180°. ∠CDB=135°. ∠B+∠CDA+∠DCB=180°[Angle sum property of a triangle]. Then if we construct a similar congruent triangle to CAD say CAE. we get x+x+15°+45°+45°=180°[angle sum property of a triangle]. we get the value of x as 45°. So, this was a more easy technique😀😀
Хобби
Sir, beautifully & patiently explained. You have an ocean of patience which is the prime requirement of any teacher
Even a student weak in Maths will easily understand if u take a class
Thanks
Wow!
Thank you for your nice feedback! Cheers!
You are awesome Pracash😀
@@PreMath I like the way you quote a theorem for every step.
You can use tringonometry too
Then ,It will be so short
It is very long method
I think you should work on more specified way to solve problem
I was drawing angles after angles, but no joy. Your explanation is great! Thanks!
Sir, you are truly the Sherlock Holmes of trigonometry. You assemble the evidence point by point, and arrive at the solution step by step. It is making mathematics good fun! Thank you.
This is geometry. Simple euclidean theorems taught in 6th and 7th. Not trigonometry.
Explained thoroughly well. Nice illustration
very nice - designating point "P" was critical, and unfortunately I didn´t see that on my own - jajaja
So nice of you Charles.
Thank you for your feedback! Cheers!
You are awesome.
Keep rocking😀
I reach the step to use the equal length. But how? So hard..
@@kwccoin3115 Try backing up and muscling it through with law of sines and law of cosines.
Yeah
Finding out what construction to make is always the hardest part ... I can never figure it out.
Felt like a suspense story with a great payoff at the end! Well done good sir.
Nice geometric solution!
I got there with tangents function. If you complete the right-angled triangle in the bottom left, with height h and base p+2d (were d is the length AD) then you see there are three right-angled triangles and by considering each one in turn that:
tan 60 = (p+2d)/h = sqrt(3)
tan 45 = (p+d)/h = 1
tan (45-x) = p/h
Multiply both sides of the second equation by two and subtract the first equation from it and you have 2-sqrt(3) = p/h
Substitute this into the third equation and you get:
x = 45 - tan-1 (2-sqrt(3))
which is 30 degrees.
Correct
Let's say that you are required to solve by construction only, no trigonometry. Eastern Brown's solution adapts as follows. Complete the right triangle and make use of the known ratios of sides for the 60°-30°-90°, 45°-45°-90°, and 75°-15°-90° right triangles. The smallest right triangle has p/h = (2-√3) = (√3-1)/(√3+1) which matches the 75°-15°-90° right triangle. The smallest angle, 15°, needs to be subtracted from 45° to get x = 30°.
superb
👍👍👍
Great job demonstrating how one can use simple facts of geometry and a little deduction to reduce the problem to a simple math problem.
In 1970, my Maths teacher at Whitchurch high school in Cardiff, Mr Smth, was so good he helped me to learn how to do this stuff and I am forever grateful. I still haven't found a use for the integration of 1 + Tan squared (x) though. :)
Great puzzle- I needed the DP portion to start before I could solve it. Thanks for providing such good content and a mix of easier and more difficult problems.
Glad to hear that!
Thank you for your nice feedback! Cheers!
You are awesome Jay.😀
Love and prayers from the USA!
None of that was DP math...
nice one!
I like the ones when you need to make an extra lines in order to solve.
After watching few similar videos able to solve this one in mind. It's mostly in finding/drawing the correct isosceles, equilateral or congruent triangles with the equal length sides.
Another flash of brilliance from the maestro, that was a superb demonstration, I'd never have got that, I like the way in many of your marvels that you add cunning lines that reveal the pathway, it's a joy to watch. Thanks again 👍🏻
You explained way better than my last years Math teacher did! Thanks for helping me clearly understand this process!
You're very welcome Aakash
Thank you for your nice feedback! Cheers!
You are awesome.😀
@@PreMath Thanks.
I am 72 and a retired electrical engineer. I guessed it correctly however, your step by step explanation was excellent. Thank you.
Don't know what your age or former occupational title has to do with it. And you "guessed" it correctly? There's no guessing as you couldn't possibly know if you "guess" was correct until someone else DEMOSTRATED it to be correct. Ugh You must have been an excellent engineer. SMH
Same. It was fu. Reliving the old steps
I haven't played with this stuff for 60 years but that was beautiful, man. Kudos and thank you.
Just sane... :^) Saint
Благодарю за разбор интересной задачи.
Well done, I didn't think it was possible to solve this problem without calculating a single length!
You can absolutely solve just based upon sum of enclosed angles without the unnecessary analytic geometry
You cant calculate a single length. There isn't enough info.
@@anonymousman1282 true, but ratio of length is all you would need, if you would want to do it another way.
Thanks for the excellent explanation. I stared at this problem for several minutes before coming up with the answer. It seemed obvious to me that angle X should be equal to angle ABC and ABC was obviously 30. based on the given angles, but while I could prove angle ABC couldn't figure out how to prove angle X.
ruclips.net/video/Z67zK4B6cfU/видео.html😊😊
Before these things used to be a headache for me in class. I did it cause I was forced to and did the minimum.
Now it’s my entertainment and I can’t get enough of it. I rather watch this than a series.
Beautifully clear.
Interesting solution. Another way is drawing CH and proving that A and H are one point, because the other two possible positions of H lead to contradictions with the triangle CHB.
Thanks a lot sir , i spent too much time thinking how to solve this and got to learn so much , indeed it was a wonderful experience.
Very beatiful questions! Thank you so much!
Wow, such a question from basic 'looking' shape. Nice job sir
Beautiful solution to the problem. Very much in the domain where math is art itself. I have opted for a cruder way (PS: I am an engineer). I dropped a perpendicular instead and used expressions for tan. Got two equations and soleved simultaneously to get the answer.
Kudos to anyone who can solve problems like this, but my hat is really off to those who can CREATE problems like this.
It's over 30 years since I took geometry in university (a Euclidian and non-Euclidian course). I thought it was not possible to find X. I also misinterpreted the double hashes as meaning AD and DB were parallel :D So I didn't get far. I was amazing by the explanation. Such a smart tactic. Thanks for taking the time.
Please make this kind of geometry video more...and make pdf including all their rules and strategy
I used the Law of Sines and Cosines. It was simpler that way (for me anyway). Thanks.
I used the law of sines and cosines. Since line AD = DB it's easier to make an assumption about the length and then we can workout CD using the sine rule and CA using the cosine rule then we can use the sine rule again to find the value of X. But thank you, this was a good exercise.
Yes - sine rule - that is how I would have done it.
Exactly.. and using this method, you can find that 2x = 60 in less than 1.5 minutes. That's way more than the allotted time for solving a geometry problem like this in CAT exam. We usually have 25-30 seconds for solving such problems in CAT exam.
How you did it? I used cosine and sine law but took like way more time than 1.5min
I had no clue how to start the first step. Very nice and beautiful. Point p is strategical
I solved by using trignometry and the law of sines for triangles. With reference to your diagram, let AD=BD= a, and let us designate the length AC=b. Then, for the triangle ACD,
a/sinx = b/sin45 ==> sinx = a/(b*sqrt 2). Also for the triangle ACB, 2a/sin(x+15) = b/sin 30
==> sin(x+15) = a/b = sqrt 2*sinx. Therefore, sinx*cos15+cosx*sin15 = sqrt 2*sinx. If we divide this equation by sinx, it can be reduced to cos15 + cotx*sin15 = sqrt 2 ==> cotx =
(sqrt 2 - cos15)/sin15, or tan x = sin15/(sqrt 2 - cos15). The right hand expression can be calculated as 0.57735026919, which happens to be the tan 30 degrees.
Very good
I solve it same way ❤️❤️
True, but the video shows the fundamental or the manual way ehe, even for those who haven't learned trigonometry still can solve the problem... Nice video ☺️
Except that trigonometry applies only to right triangle.
@@dilasgrau6433 Nope. Law of sines and law of cosines is also considered trigonometry. Anything that uses trig ratios is considered trigonometry.
Very thorough explanation. Saw similar videos and thanks for posting those. was wondering if there a pattern to solve such problems?
i guess in these kind of problems only the one who created them can solve them, or you can go through a long journey of trying all possible methods and theoremes to finally conclude it
Thank you, very well explained, i appreciate the attention to detail in the explanation.
Another fascinating explanation Man !
Thankyou Sir for the excellent and clear explanation. I succeeded to solve using the sin theoren (but it takes a lot of time).
😂
In ∆CDB, angle B + 15° = 45° [exterior angle is the sum of interior opposite angles]
Therefore angle B = 30°
Let AD = DB = a
and AC = b
In ∆ABC by law of sine
2a/sin(x+15°) = b/sin30°
= b/(1/2)
= 2b
=> sin (x+15°) = a/b --------(1)
In ∆ ADC
a/sin x = b/sin45°
= b/(1/√2)
= b√2
=> √2 sin x = a/b ---------(2)
From (1) and (2)
√2 sin x = sin (x+15°)
√2 sin x = sin x cos 15° + cos x sin 15°
(√2-cos 15°) sin x = cos x sin15°
sin x / cos x = sin 15° /(√2-cos 15°)
tan x = sin 15° / (√2-cos 15°)
tan x = 1/√3 ( pl. refer Note below)
= tan 30°
Thus the unknown angle x = 30°
Note:
sin 15° = sin (45°-30°)
= sin 45° cos 30° - cos 45° sin30°
= [1/√2] [ √3/2 - 1/2]
= (√3-1)/2√2
Similarly
cos 15 = (√3+1)/2√2
√2-cos 15 = √2 - (√3+1)/2√2
= (3-√3)/2√2
= √3(√3-1)/2√2
= √3 sin 15
Hence
sin 15° / (√2-cos 15°) = 1/√3
You explain this so much better than my teachers did in high school.
That was a nice problem! Thanks for the video!
an easier solution i think would be to just draw a line say CE parallel to AD now using alternate interior angle property we can conclude that angle ADC = angle DCE. Now since the line CE is also parallel to DB therefore we can conclude using alternate angle property that angle DBC = angle BCE. Now angle DCE= angle DCB + angle BCE = 45. Therefore, 15 + BCE = 45, therefore BCE = DBC = 30
thats not x
I admit I got stuck after step one. designating point P was a clever trick, but only worked out by chance, due to the particular properties of the two triangles. After it turned out that the vertex in D of APD is 60 degrees, the rest was easy!
No worries Philip👍 This was a challenging one indeed!
You are awesome. Keep persevering😀
Excellent explanation . Thank you .
Amazing solution!! Congratulations.
Intuitively, I figured that the bifurcation of the base created a triangle ADC which you could map to ABC by a reflection across AC, and a rotation around A, followed by a dilation. This would show that angle BAC is the same angle as CAD. The measure of angle BAC I got deriving angle DBC from 180 - 45 = 135, and then 180 - 15 - 135 = 30. As it turns out, this leads to the correct solution, namely x = 30 degrees. But I don't know if this is a fluke :)
Not a fluke. The method presented os a very long winded approach to achieve the same result.
3:13 to get angle PDC you don't need the 120 degree. Just use the exterior angle theorem again in the trangle PDC with angle BPD=30degree as the exterior angle, and angle PCD=15degree so angle PDC=30-15=15degree.
This is a good question that looks difficult but once the additional lines are drawn it becomes straightforward.
Thanks!Wish you happy new year!
I felt like you taught me something and I understood it. That equals to having learned something. Thank you
The hardest part of solving geometry problem is to find a point like P.😂😂😂
I agree.
Gosto muito de seus ensinamentos, obrigada!
De nada, Sonia!
Obrigado! Saúde!
Continue agitando😀
Amor e orações dos EUA!
great explanation sir.. really appreciate your work
Excellent explanation, thx for your effort.
Yes, I found “X”. It was hiding just to the lower right of “C” at the top. It was pretty easy to find. Especially since it was written in red! 😊 Just kidding. I never thought I would use calculus or trigonometry much when I was in high school. Then I got into machining and use it almost daily! Pay attention kids this IS important!
Hey, can I ask more about the field in which you work?
@@Manifestingqueen1111 sure
Well, there are many ways of solving this beautiful problem!!
1) using elementary geometry.
2) using sine rule
3) using m-n theorem
Using the third method gives u the answer in just 1step...😎😎😉😉
Anyways, thanks for sharing!
Love from India!!
M-N theorem is very handy and interesting. Went through its proof.
(m+n)cot(theta)=m.cot(alpha)-n.cot(beta).
In this problem
AD=m DB=n; m=n
Theta=135
Alpha = X
Beta = 15
Sin(15) = (√3-1)/(2.√2)
Cos(15) = (√3+1)/(2.√2)
CotX = √3
X = 30.
Very nice approach.
Today i learnt m-n theorem.
Thank you Kusuma
I used the second approach
@@sandanadurair5862
Nice 👍!!
Thanks for sharing! Cheers!
You are awesome Kusuma😀
Love and prayers from the USA!
When do you use the m-n theorem?
How to solve using m-n theorem ?
Good explanation.
Thanks for your sharing.
This seems to be one of a number of similar problems that are special situations. We are given a 45 degree ray from the midpoint D of AB and an intersecting ray from A to C. The problem is to find angle ACD. The method of solution only works if the point C is chosen so that angle DCB is 15 degrees or alternatively angle DBC is 30 degrees..
I was just wondering how will you use that information, that those two parts of bottom line is equal, and I'm speechless how beautifully you created a triangle out of it and used that fact... I knew that equal sides of triangle subtend equal opposite angles in a triangle, but i wasn't smart enough to use that fact... Thank you sir, love from India..!
Given CD bisects side AB in equal parts, it would also mean that CD bisects angle C in equal parts. Let me know your thoughts incase you feel I am wrong and why, happy to be corrected.
I used the law of sines to get to the same answer. It's a bit more trig identity fiddling but less subdivision and added triangles.
Very nice approach. I solved it using sine rule. sin(15+x)/2sinx=sin30/sin45=>sin(15+x)/sinx=sin45/sin30. Putting x=30 or 15+x=45 gives the same result x=30. Hence x=30 deg.
Great Satyanarayan
Thank you for your feedback! Cheers!
You are awesome.😀
Trigonometry works only on right angled triangle, right?
*Trigonometric solution:*
Drop a perpendicular from C to meet the base line at E. Let CE= a and ∠ECA= θ. Since ∠DCE=45°, we have ED=a, and since ∠EBC= 30°, we have EB=a✓3. This gives AD= (✓3-1)a and EA=ED-AD=(2-√3)a.
Now Tanθ=EA/CE = 2-√3, which gives us θ=15°. Hence X=30°.
Great!
Thank you for your nice feedback! Cheers!
You are awesome Hari.😀
Very well thought out. Really nice alternative solution
Clearing away the cobwebs. 55+ years I was solving problems like this as part of a submarine fire-control problem (I.E. what angle to set the torpedo to intersect with the surface target) We had a mechanical analog computer, however, if it failed, you had to know how to do it by hand. Thank you. Narragansett Bay
So simply explained.... thanks
I got it. My sister told me the triangle CDP is an isosceles because angle PCD and PDC have the same 15 degree. Therefore, PC = PD. The triangle APD is equilateral, hence PC = PA. So angle PCA and PAC are 45 degree each. Consequently, angle ACD is 30 degree. That’s awesome!!! Thank you very much.
I had learned geometry 50 years ago. I have to confess this brings back a lot of memories.
Excellent! I learn geometry and English language in this channel! Thanks!
You're very welcome Sergio!
Glad to hear that!
Thank you for your nice feedback! Cheers!
You are awesome.😀 Keep it up!
Love and prayers from the USA!
A class one teacher and a very challenging geometry problem.
You keep up the name Indians are great mathematicis
What software do you use to make these geometric shapes?
I couldn't solve it thanks for this video 👍❤️
No worries Milli 👍
You are awesome. Keep persevering😀
Great puzzle! I didn't find this excellent geometry solution. Instead I dropped a perpendicular from C to point E, then used trig to get CE, EA and EB (after setting AD and BD=1). Then I used Pythagoras to get CB=sqrt 2. Then the Law of Sines to get sin x=1/2. Thanks PreMath!
Glad to hear that!
Thank you for your feedback! Cheers!
You are awesome William😀
How did you fin CE ?
Could you please explain
@@hussainfawzer Hello Thanks for the question, I left out a lot of steps and also erred in saying CB=sqrt2. Sorry for that (CA=sqrt2 is correct).
To get EA then CE let EA=y and AB=1 (or any number but 1 makes the calculating easier)
Since triangle CED is a right triangle with a 45 degree angle, CE=ED=1+y. EB also =2+y
Since angle CEB is defined as a right angle and CBE (as shown by PreMath) is 30 degrees, CEB is a 30-60-90 triangle and CB will be twice CE or 2+2y
So by Pythagoras CE^2+EB^2=CB^2 or (1+y)^2+(2+y)^2=(2+2y)^2
Simplify and get 2y^2+2y-1=0 and so y=(sqrt3-1)/2 and EC=(sqrt3+1)/2
I hope that answers your question, Hussain. Now one can get CA=sqrt2 from Pythagoras. And looking at triangle CAD, the law of sines gives sinx/1=sin45/sqrt2 and sinx=1/2.
@@waheisel
That was beautiful math. It reminded me of a lot of stuff. Ty
I used to love math and physics with a dream to be get into cosmology. Due to personal reasons I'm now soon becoming a doctor.
It's good but I miss math and physics so badly 😅
@@yuda4626 Thanks for the kind comment Ty. When I was in school I also liked math, physics, and astronomy. I wasn't nearly good enough to make any of those a career. I also became a doctor. Even though you won't use so much math and physics in your career you can still enjoy your daily PreMath puzzle! Best Wishes
Aoa. Very well explained in very simple and clear language. Enjoyed it really. 👍👍👍
That was a long process, but I got it right with a shorter process. The last time I took math was 60 years ago. I had an excellent teacher!
I used the sine rule twice and the sine addition angle formula, but this solution is far better.
Thank you for your feedback! Cheers!
You are awesome Barry.😀
THE SHORT:
If you could draw DP so that DP=DB, on paper, you could also just as easily draw AP so that AP=AD=CP=PD, on paper... all anchored by the given angles of ADC & DCB.
THE LONG:
Imagine vertices ABC are 3 stars in a system, with D being a star exactly between star A and star B (let's say with Hubble, we're easily able to measure that the star D is exactly half way, in light years, between stars A & B, all three nicely aligned).
From ANY of these given stars, to any other star (in the ABCD set of star system), as well as, angles 15 & 45....nothing was given, could have been given, via assumption or "eye balling".... Everything came about through precise measurements via parallax + trigonometric, standard Candle method, etc.
Take one step back and ask yourself: --- what is the astronomical (if we took ABCD to be a set of 4 stars in the Milky Way) or geometric logic that said you could, for example, draw a precisely KNOWN line segment, like DP, from one vertex of BDC, to the opposite side, BC, in such a way that DP is exactly equal to one side, DB.... but NOT be able to do a similar thing, say, AP?
After all, AP=AD=DB, DB=DP, and AD=DB.... all anchored by the given angles of 15 (DCB) & 45 (ADC). For example, "P" can NOT be in any other place, along BC, such that AD is not equal to AP (due to the dictates of the angles 15 and 45, as given).
In other words, there's neither "wiggle room" for angles ADC & DCB, on the one hand, nor for point "P" along BC such that AP=AD=DB=DP is NOT true, on the other.
Man you're fascinating! Things like this never comes to the mind.
非常に難しい問題でしたが、何とか自力で解けました！
I made it !! Thanks a lot!
素晴らしい仕事😊
ありがとうございます！乾杯！
あなたは素晴らしいです。続けてください😀
Love and prayers from the USA
I learned from you, thanks!
During my jee preparation (engineering entrance exam of India) i studied a theorem called m-n cot theorem, and really it gave the answer in just 5-10 seconds :) 😁
BTW great explanation sir😊
@Arnab Karmakar XI SC 6 Good job bro !!!All the very best 😀😀
Excellent little problem, thank you!
Can we solved simpler, I think, using law of sines.
In triangle BCD, sin
Cool! Many ways to solve this problem.
Thank you for your nice feedback! Cheers!
You are awesome Vsevolod.😀
I use the same way to slove this question.
can u please plz tell me only what does sin(135-x) equal? i didn't get it from ur comment
@@Z7youtube It's in my comment. Using formula for sine of a difference.
sin(135°-x) = sin(135°)cos(x)-cos(135°)sin(x) = (√2/2) cos(x) + (√2/2)sin(x)
We used the fact sin(135°)=sin(45°)=√2/2 and cos(135°)=-cos(45°)=-√2/2
@@vsevolodtokarev oh ok i didn't know about the formula for sine of a difference , tysm!
teacher I think you taught is very well, and made my ideas very clear, especially my observations.
Very clever problem. I thought you would need trig so I gave up on it. Very impressive that you can solve this without needing to use trig.
There are many ways to solve this kind of problem!
Thank you for your nice feedback! Cheers!
You are awesome Lance😀
could use trig by drawing a perpendicular line from C to P to form right triangle CPA. Triangle CPB is a 30-60-90 right triangle. label the sides as 1, sqrt 3 (or 1.732) and 2 let's label angle CPA as 'y' . y + x =45 degrees and you now have and 45-90-45 with sides 1, 1, and sqrt 2
degree right-triangle line PD =1 and CD= sqrt 2 . Using ASA (15 degree, sqrt 2 and 135 degree) for triangle CDB gives 0.732 for DB , but DB = AD thus PA = 1-0.732 = 0.268. Since the right triangle CPA has the sides 1, 0.268 then using SAS (1, 0.268, and 90 degree) gives 15 degrees for angle y,
but y+x=45 degree hence x=30 degree
Is it only me who thought that if AD=DB, CD is the median line of the triangle in C, thus meaning in cuts the angle in half? And if DCB is 15, ACB would be 30?🤔🤔
You’re correct especially if you check with Sine law
Actually, that is not always the case, as shown in this problem here. The median of a triangle in a given vertex is not always equal to the bisector of the relative angle. They're the same in some cases, for example if the triangle is equilateral, it is true for alla the vertices.
@TJ the sine law is applied within a given triangle, in this case you can say tha
sin(15°)/(DB) = sin(30°)/DC
And
sin(x)/AD=sin(135-x)/DC
You're mistaken between median line and angle bisector.
This is proven false in this example so if it cit ACB in middle = ACD = DCB =15° which is not the case
Hi, you could have separated triangle cdb and joined AD and DB by rotating clock wise triangle CDB. Making points A and B to be the same point. And you can get the answer in 2 steps. As rotating would create an isosceles triangle. Try it out
Rotating does not create an isosceles triangle. In fact, in the general case, the rotation is more likely to create a 4 sided polygon. In this particular case it creates a triangle, but not an isosceles one. To think this through, think about point C being split by the rotation, where C1 is the original point C, and C2 is created by the rotation. In this case, because angle ADC is 45 degrees and angle CDB is 135 degrees, after the rotation the line segments C1D and DC2 are collinear forming a straight line, with D in the center, since C1D and DC2 are the same length (but since they form a straight line they are not two sides of an isosceles triangle). So one side of the resulting triangle is C1C2, with the other two sides being AC1 and BC2 (A and B are the same point after the rotation). One of the angles in this new triangle is the 15 degrees from angle DCB prior to rotation, another angle is X degrees, formed from angle ACD prior to rotation (now AC1C2 after rotation). We know from the solution presented that X is not 15 degrees but is a value (trying not to spoil) such that none of the angles in the triangle formed by the rotation are the same.
Damn!! Very good explanation!!
Excellent. Thank you
Fantastic and clever solution! I couldn't made it that way!
I have another aprroach:
1. Draw the circle that contains point A,B,C. To do so, draw 2 perpendicular lines at the midpoind of AB and BC, these two lines meet at the center O. WE notice that AOB is the angle at the center and the angle ACB (= angle x +15 degrees) is the angle at the circumference.
2. Demonstrate that the angle at the center AOB is equal 90 degrees. Then, we can find the value of the angle ACB = 1/2 AOB= 45 degrees.
3. The angle x= angle ACB - 15 = 45-15= 30 degrees.
You cannot say that there is a circle which contains point A,B,C
@@emaceferli726
Yes, I can. Just label M and N the midpoint of AB and then AC. Draw 2 perpendicular lines to AB, and AC at these points which meet at the center O. Then we have OC=OA=OB, therefore we can draw a circle from the center O with the radius R=OC=OA=OB.
It is easy to see that the angle CBA=30 degrees (CMA is the exterior angle=45degrees).
The angle CBA is the angle at circumference of which COA is the angle at the center, so COA= 2 CBA=60 degrees----> the triangle COA is an equilateral one.
Let's say I is the midpoint of OA. The height from the vertex C to I is also the bisector of the angle ACO (=60 degrees).
Because the angle AMC=OMC=45 degrees so CM is the bisector of the angle OMA, so C,I,M are colleniar.
Thus the angle x= ACM= 30 degrees
Trigonometry makes the whole thing much easier. Actual challenge is to solve it without trigonometry. ☺️
Nice, challenging problem. It took me a few minutes to solve the problem in my head!
My question is, when we create line DP = DB, how do you prove that point P were located on line CB?
I took Geo Trig last school year and felt very smart knowing how to solve everything. However when I saw this, it was not the case.
Thanks for explaining this very thoroughly. It was very fun and interesting to learn how to solve this Geo Challenge. Please make more!
I have failed to solve it.🙁
No worries 👍 This was a challenging one indeed!
You are awesome Mustafiz. Keep persevering😀
So crystal exoplanation. Amazing
Waiting another
Notice that sin(x)/AD = sin(45)/AC and sin(x+15)/(2*AD) = sin(30)/AC. Combine to get 2*sin(x)/sin(x+15) = sin(45)/sin(30). Upon expansion of sin(x+15) and some algebra, we get x = 30 degrees.
I solved it using sine rule, without any construction. Good problem !
How pls tell me
Great Subramaniam dear.
Thank you for your nice feedback! Cheers!
You are awesome.
Keep rocking😀
I used the same method as well! Law of sines to get it
So difficult
This problem gave me a lot to think about
No worries 👍
You are awesome. Keep persevering😀
Really nice. Excellent. You did it by simple geometry - fantastic solution . Thank you..
Glad to hear that!
Thank you for your nice feedback! Cheers!
You are awesome Debdas dear.😀
Love and prayers from the USA!
Thank u sir for ur knowledge 🙏
I think the question may be solved by applying basic triangle rules. I tried by following method-
1. By external angle theorem: angle DBC = 30°
2. In triangle DBC, if length of DB is 'a' (against 15°), then CD is 2a (against 30°)
3. Now, in triangle CAD, AD = a, CD = 2a
4. Angle ACD = x, Angle ADC = 45° = y, Angle CAD = z
5. Angle z = 2 (Angle x) [because CD = 2 (AD)].
6. As, Angle x + y + z = 180°
x + 45 + 2(x) = 180°
3x = 180° - 45° = 135°
x = 135°/3
x = 45°
Cool solution
But x = 30°
angle double doesnt mean side is double...its a sine relation
There is no rule in a triangle as you wrote "if length of DB is 'a' (against 15°), then CD is 2a (against 30°)". By this info you can just write: CD>AB.
Honesty i couldn't solve it but very good question to practice thanks😊
No worries Pranav dear 👍
You are awesome. Keep persevering😀
@@PreMath thanku
As a first step use the 'Thales' theorem to get the 90 deg angle at P. Everything else follows automatically.
Can you tell me which app or software is it on which you made writing video
It was an easy puzzle! You could've solved it with a much easy technique. ∠CDA + ∠CDB=180°[Linear Pair] . So, ∠CDA is given as 45°. Then, 45°+ ∠CDB=180°. ∠CDB=135°. ∠B+∠CDA+∠DCB=180°[Angle sum property of a triangle]. Then if we construct a similar congruent triangle to CAD say CAE. we get x+x+15°+45°+45°=180°[angle sum property of a triangle]. we get the value of x as 45°. So, this was a more easy technique😀😀
Easy approach
That's the one I was thinking of aswell.
But the value of x is 30°, it says.