Find Angle X in this Compound Shape | Step-by-Step Tutorial
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- Опубликовано: 7 авг 2024
- Learn how to find the unknown angle in a compound shape made up of circle and a triangle by using the exterior angle theorem. Quick and simple explanation by PreMath.com
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Thnx for such brilliant logical question
So nice of you Pranav
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You are awesome 😀
I didn’t know the trick with the angles of the intercepting arcs, but once explained, it’s as logical and as straightforward as can be.
Angle substended by arc at centre and circumference par nhi banega
I love your thumbnails. They make me guess solutions before I click your thumbnail.
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Well, the angle opposite of 95 is also 95. The four sided figure with the angle of 40 has a total of 360 degrees. Since the two unknown angles are identical,
x = 180- (360-95-40)/2 and that is 67,5 degrees
Very well done.
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You are awesome Henk😀
Yes, that's the shortest method. Solved it mentally in 1 minute.
Much easier to solve this way.
Very short approach
Yeah, that’s a good method. But, I wouldn’t hav3 learned about intercepting angles on an arc. Now I know both. Thanks Henk.
Thank you for video. I did it by different way. The angle opposite to 40 is 95, therefore with the quadrilstetal, both angle touching the circle is (360 - 95 - 40)/2 = 112.5.
X is adjacent angle of 112.5, so X = 180 - 112.5 = 67.5
Excellent!
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You are awesome Paul😀
Reading the comments, I'm impressed with all the many ways to do this problem! Everyone give yourselves a hand. :-)
x+α=95, 40+α=x.
40+α=x → x-α=40.
Thus, 2x=95+40=135.
The value of x is 67.5.
Super
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I did it this way too.... took a few minutes to spot I'm not embarrassed to say.... :)
I solved this before watching with a simpler, yet, more limited approach. This method will serve superior to my previous one in 3D applications. Thanks for sharing this.
The exterior 40 degree angle is one half the difference between the major and minor arc it intercepts. The 95 degree angle is one half the sum of the arcs the chords creating it intercept. From that you can easily calculate the major arc to be 135 degrees. X is just the inscribed angle which is half of 135 degrees, therefore 67.5 degrees.
27,5
Sir you are a living legend! Love from Pakistan 😊👍🌹
Thank you Sameer dear!
You are the best. Stay blessed😀
О! К сумме углов треугольника добавилось сумма углов четырёхугольника! Гениально!!!
thanks to your videos, I tried to solve this by myself and i got it right. I hope i don't need to struggle in class 7 next year. Also have a good day.
Excellent. I got the same answer with a very different method. Where the lines cross within the circle you have two pairs of angles at 95 and 85 degrees (one of the 95's is given). The right hand 95 degree angle is one angle of a kite, and the 40 degree angle is the right end angle of the kite. 95 + 40 = 135. Angles of the kite add up to 360 degrees, so the two remaining angles are each 112.5 degrees ((360-135)/2). One of these is an angle on a straight line with x, so 180 - 112.5 = 67.5=x.
Thank you for a nice geometry question.
With respect, x=α+40⇒α=x-40
But x+α=95
⇒x+x-40=95, 2x=95+40=135
x=135/2=67.5°
Great job
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You are awesome Hassan😀
There is actually a pretty easy solution. If you connect the midpoint and the point outside the circle you can use two times the exterior angle theorem to find that 2x=40+95 degrees.
Also we can use the degree of the arcs to solve it. If we set the big arc on the left side to be A and the small arc on the right side to be B then we have the 40 degree angle equals (A-B)/2 while the 95 degree angle equals (A+B)/2 from basic arc and angle relationship knowledge. Then we can easily solve A=40+95 degrees while x=A/2.
this was a very easy question even for those who have not prepareed for IOQM
x = (40+95) / 2 = 67.5
because angle between black lines, 40 = +x -95 +x , where -95 is vertically opposite to the given 95 ; and x is in both triangles as these are similar triangles.
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You are awesome David😀
I think this is the best way to start angle chasing geometry on olympiad lvl
Good form dictates writing the =signs in a straight line.
Other angle on circle that share the same arc of 95° and 40° is also X.
Complimentary angles of X are both X' -> X'=180°-X -> X=180°-X'.
Opposite of 95° is also 95°.
Quadrilateral's sum of angles is 360°.
->
360°=95°+40°+2X' -> 360°=135°+2X' -> 2X'=225° -> X'=112.5° -> X=180°-X' -> X=180-112.5° -> X=67.5°.
Nicely done!
Thank you very much.
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Sum of angles of quadrilateral = 360 and the same on a line = 180. Therefore, 40 + 95 + 2 x (180 - x) = 360. Therefore, x = (95 + 40) / 2 = 67.5 deg. *Easy. Right* ?
Draw a Line opp 40 deg angle joining the 2 points on the circle vertically. So this will form two triangles one having 40 deg angle. Since these 2 triangles are isosceles triangles so the 2 angles on 40 deg triangle will be 70 deg each. The other triangle will have 95 deg. as one angle the other 2 angles will be 42.5 each ( 2*42.5 + 95 = 180 ) So angle on the other side of line of X will be 42.5+70 = 112.5 so X = 67.5
So, haven't watched this yet, but my intuitive approach:
Imagine a line that bisects both the 40° and the 95° angles. The 2 lines that make the angle x are then 95/2° and 40/2° on either side of this bisecting line.
95/2° +40/2° = 47.5° + 20° = 67.5°.
You should have add that the circle exterior angle is equal big chord minus small chords. So, after you have found alpha chord as 27.5, the X chord is then 40 = X - A, i.e. X = A+40 = 67.5
Angle substended by arc at centre and circumference par nhi banega
It's pretty straightforward: x+alpha=95 and x=alpha+40. Solving for x yields x=(95+40)/2=135/2=65.5
What is the name of the program that is used to write like this and the solution?
I drew line between bottom two points, then the angles of the black and red triangles are 70 and 42.5 leaving 27.5 for alpha, then x=180-27.7-85 = 67.5 since angle opposite x is 180-95,
That only works if those triangles are isosceles, which you can't guarantee. However, you can say that the base angles of those 2 triangles sum to 140 and 85, and add to 225, which must be the same as the sum of the base angles of the other 2 triangles on that base, each of whose apex angle is x, so 2x + 225 = 2 × 180, which gives the correct answer.
Sir u r lord Shiva for us. Kindly carry on all ur superb problems. Take care of ur health. If possible then kindly present us some tricky geometry problems of K.C.Nag.
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You are awesome Ram. Stay blessed 😀
Much simpler way:
Opposite angles of intersecting lines are congruent, so 95 degrees on both sides.
Extend line from intersection to apex of outside 40 degree angle. Measurements are 1/2 of before, so 47.5 and 20 degrees.
Add 20 to 47.5 to get x=67.5
To get the rest: Subtract 20 from 47.5 to get alpha=27.5
Supplementary angle of 95 is 85, so beta angle of triangle is 85.
A triangle has 180 degrees, so you can check by adding 85+47.5+27.5=180.
Excelent
Thank you Susen😀
The Sum of the Quadrilateral will be (180-X),+ (180-X)+ 95 + 40 = 360
hence 2X= 135
hence X = 67.5
By the way, my comment is just to show another method. Science, in every form or shape, is just fun!! Nothing against premath, on the contrary!!
Great point!
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You are awesome Ken.😀
Alternative solution:
The two red lines form four angles. 95° to the left, so also 95° on the right. That leaves 170° for up and down, so 85° each.
The tip of arrowhead shape bounded by the red lines forms a diamond shape with the angles 40° at the tip and 95° at the opposite end. A four-sided shape has internal angles of 360°. Minus 40° minus 95° leaves us with 225° for the top and bottom angle, so 112.5° each.
Now we have two angles for the triangle formed by the alpha angle and the 40° angle, which are 40° and 112.5°. A triangle has internal angles of 180. Minus 40° minus 112.5° leaves 27.5° for alpha.
As the final step, we know that the bottom angle of the two crossed red lines is 85°, and also that alpha is 27.5° Together that makes 112.5°. Subtract that from the 180° for a triangle, and that leaves x with 67.5°.
The weakness of this approach is that it assumes the whole shape is symmetrical, but does not prove it. If it were not the case, I think the solution would not work.
40 is equal to half of big arc minus half of small arc. x is half o f big arc, a is half of small arc so x-a = 40, x+a = 95 => 2x = 135 => x = 67.5
12 sec for this one...for calculations.(+,-) Have a great day! and merry Christmas!!!
#inscribedangle #congruentangles #intercept
#exteriorangle #exteriorangletheorem
easier than I thought 👍😂
Bravo
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You are awesome Milli.😀
Please sir I want you solve this question if (m+n)^2=m^2+n^2, what is the value of (3^m)^n you have contributed great to improve myself may Allah elevate higher Ameen
Answer would be 1 is suppose.
x=a+40.,,95=a+x., from both a is removed my subtracting one equation from the other then 2x= 95 +40, therefore x = 135/2,=67.5
I have another solution on this problem:
First of all I rewrite this problem as,
Let A,B,C,D are concyclic;AC and BD intersects at point P and AB and CD intersects at point Q,such that
I solved it by equation x = 95° - x° + 40° (Ext. Angle Property)
Oh, this is not a contest. I see comments like “I did it in 40 seconds”.
For me, Its just to show there are other ways to get to a solution. Nothing more. Lets not get carried away. Just look at this as a challenge and give your solutions. Nobody is better, nobody is worse.
Big fan sir
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You are awesome Anita Joshi. Stay blessed 😀
Love and prayers from the USA!
95°=(arc x + arc a)/2 ; 40°=(arc x - arc a)/2; arc x = 135° -> x° = arc x /2 = 135°/2 = 67.5°
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You are awesome Mihai.😀
I solved this equation with another method :
We have :
180° - x + 40 + 180° - ( 85° + x ) = 180°
So :
315° - x - x = 180°
- 2x = 180° - 315°
- 2x = - 135
x = - 135 / - 2
x = 67.5°
Done
67,5°
Once you know alpha 27.5°, x=27.5+40 =67.5°(exterior angle theorem)
Very nice
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You are awesome Mahalakshmi dear😀
67.5 degree
Super Nihal
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#sum #anglesum #sumofangles
67.5 °
Super Mumtaz👍
Using exterior angle property
X= alpha+40. ....1
X+alpha = 95.....2
From 1
X-alpha. =40.....3
Adding2 &3
2x = 135
X= 67.5
Excellent Sandanadurai
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Ext angle 95=x+x-40
x=67½
one way another way
from the fig 0:01 x=y+40 and x+y=95
x=67.5
I couldn’t solve it cos I duno the fact about the angles in inscribed angles
No worries Garrick. We are all lifelong learners
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@@xxrgxxcasco don’t worry. All is in good taste. I agree sometimes I just don’t see it, but I still do enjoy problem solving. Sometimes I do feel the solution presented was more confusing by my own solution lol.
1)95°+40°= 135°; 2)135°/2=67,5°✓
Excellent Ihor
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because.... 95+(reflex angle 265)= 2theta angle + 40 +( reflex angle 265)=360.....therefore 360 -( reflex 265)=2theta angle+40=95.....therefore theta angle =27.5.....then x=95-27.5'=67.5
x = 95 - a and x = 40 +a. Adding, we obtain 2x = 135. Thus x = 67,5. End!!!
2*(180-X)+95+40=360
360=95+(180-x)+40+(180-x)
a = x - 40 = 95 - x
x = 67.5
Super Greg
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x + alpha = 95°
x + 27.5° = 95°
x = 67.5°
Please name it as A,B,C,D in order to avoid confusion
Легко доказать чтр х среднее арифметическое углов 40 и95 градусов
I solve it by looking in the thumbnail only
alpha + alpha + 40° = 95°
2 alpha + 40° = 95°
2 alpha = 55°
alpha = 27.5°
This seems overly complex. Each of the two described triangles are isoceles with a shared base. So the black triangle has base angles of 70 degrees, and the red one 42.5. In the triangle with the angle x you have one base angle at 70 and one at 42.5. The angle x is therefore 180 -42.5 -70 = 67.5 degrees. Far simpler than the other solutions.
It's evidently not 95 degrees there.
x＋(x−40)＝95
x＝67.5
X=67,5
x+y=95; x-y = 40;
Класс!
You are awesome Jakhima.😀
@@PreMath .in fact, I'm from Russia and I don't understand English well. but there are no language barriers for mathematics, so I understand you perfectly;)
@@PreMath . and further. thanks for the detailed explanations! please continue and good luck!
x = alpha + beta + gamma
L
A
B
C P
D
E
To much easy. For school children.
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The angle labeled 95 degrees is obviously LESS than 90 degrees! Bogus illustration.