Can you Find Angle X? Step-by-step Explanation
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- Опубликовано: 11 дек 2021
- Learn how to find the unknown angle in this compound shape created by two overlapping isosceles triangles by using the Exterior Angle Theorem. Step-by-step tutorial by PreMath.com
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Superb Teacher Ji🙏
You are altogether a different level.... Coming up with good questions and solving it so that everyone understands
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Marie Van Camp is perfect. BC, BD and BA are three radii, so that a circle would pass through them with B as the centre. Now angle made by the chord AD at the circumference is half that made at the centre. So 66 divided by 2 = 33 degrees.
Simple solution. Nice
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Your comment made perfect sense except for the first sentence. Who on Earth is Mary Van Camp?
Yep, same way
@@tychophotiou6962 the one who wrote the original solution first, in the comments
A, D, and C are points on the circumference of a circle (because BA, BD, and BC are radii). That means the circle theorems come into play. The angle from a chord to the centre is twice that to a point on the circumference in the same segment, so x = 66/2 = 33.
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Thats a nice approach
Nicely done! I did it two different methods.
METHOD 1:
I assumed there is a circle going through points A, D and C. (since AB = BD = BC)
Also, point B would be the centre of such a circle.
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Plus simplement ...
Les points A, D et C sont sur un cercle de centre B.
L'angle ABD est un angle au centre de 66°
L'angle BCD intercepte le même arc de cercle AD avec son somment sur le cercle.
Sa mesure est la moitié ; donc 33°
Exactly! No nead for complicated solutions!
Super approche !
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A very elegant solution by Marie Van Camp
Très original, bravo!!!
Un angle au centre mesure le double d'un angle inscrit interceptant le même arc 👍
Mais je crois que vous vouliez dire L'angle ACD (X) intercepte le même arc AD ... et non l'angle BCD
Another way
BÂC + 66 = X + angle BDC & (angle BDC = X + BÂC) ==> 66 = 2X ==> X = 33
Nice question and nicely done. I did it your way as well as using circle theorem(angle at the centre =2xangle @ circumference. B being the centre, radii BC=BD=BA.
Angle x=66/2=33°.
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If I had teachers line you, I’d of been a mathematician! You’re a great teacher!
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very nice, PreMath!!! i did it in a simpler way:
AB=BC=>
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تمرين جميل جيد . رسم واضح مرتب . شرح واضح مرتب . شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم. تحياتنا لكم من غزة فلسطين .
The time of arabian mathematicians has gone more than 10 centuries already
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Marie absolutely correct 👍
Simple Solution
Consider Triangle ADC. Since AB = DB = CB, B is the Centre of the Circumcircle of Triangle ADC.
Since the angle subtended at the Circumference is half of the angle subtended at the Centre it follows that x = 1/2 of 66 = 33
Sumith Peiris
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Very interesting question,
fine solution finding result .
Thank you so much.
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Sir your tremendous efforts are really appreciable, love from Pakistan 😊👍🌹
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Simple problem, but your method is very so exciting.
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very well explained, thanks for sharing
I did it slightly differently, I made an additional triangle APD. This gave angle BAD as 57 and angle ADB as 57, due to being an Isoceles. Then I looked at New Triangle ACD, and the angles made the following:
57 + x + a + x + 57 - a = 180
114 + 2x = 180
2x = 66
x = 33
Simply solved by different way.
Very interesting solution.
Have a nice day.
a?
Realising that a circle with centre B passes thr' puts A,D andC is by far the best way to solve it.It is short and simple and you don't have to draw the circle to see that angleABD is twice measure of angle ACD.
Great approach!
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Man, at first I thought this task has an infinite number of solutions. Then I saw your solution and surprized. I drew this task in AutoCad for check. I rotated BC line. So, yes. Angle X is 33 degrees no matter the BC line position. Bravo!
Simply the best
The angle BAC and BCD=57°
From triangle BCD we have the angle CDB and angle BDC are equal 57+x.
Then angle CBD= 66-2x.
From triangle ABC we have 132-2x+114=180°
That means x= 33°
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Great upload as always.
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I have a simpler method! I labeled angle PBC = y. So, in triangle BDC, y+2a+2x = 180 deg. In triangle ABC, y+2a+66 = 180 deg. Equate both equations together, 2x = 66. x = 33. Good qns!
yes I am too
Great approach
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شكرا على المجهودات
نستعمل الداءرة ذات المركز Bوالشعاع BC نجد x=66/2أيx=33
Because BA=BD=BC so we can draw a circle of which point B is the center, and the radius BA, BD, BC. In this circle the angleABD is the angle at the center, and the angle x = angle ACD is the angle at the circumference so x= 1/2X 66 == 33 degrees.
Great explanation nice to watch
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Angle at center twice angle at circumference, 1step work
Great approach
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Bonjour, merci pour vos intelligents exercices de la géométrie pour les jeunes mathématiciens. Cette solution est très générale pour un exercice du genre, sauf pour celui là je propose une solution d'une seule ligne, il suffit de remarquer que le point B est le centre du cercle de rayon BA, BC et BD donc il suffit aussi de remarquer les angles x et 66° intercepte le même arc dont x =66/2 = 33°
Merci pour l'effort que vous déployer pour nos jeunes.
Inscribed angle theorem. B is the center of a circle passing through A. C and D. ACD is an inscribed angle and thus half the intersected arc AD. AD is 66 and ACD is 33.
BA=BD=BC :A,D,C on the circle of Center B.Angle X=angle ACD=1/2 angle ABD(angle at center of circle)=1/2 66=33.
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Great!
Brilliant! Another solution by the Sherlock Holmes of maths!
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Simple solution, 2 steps
BA = BC = BD so B is the centre of circle ACD
Hence = 66/2 = 33 (angle subtended at centre is double the angle subtended at the circumference)
Sumith Peiris
Moratuwa
Sri Lanka
There is actually a very easy way to solve this question although it might be very difficult to find out. Basically if you draw a line from B to segment AP that meets AP at Q, such that angle ABQ equals x, then you can easily find that triangle QBP and triangle DCP are similar due to that they have one angle facing each other, and angle BQP=angle A+x=angle BCA+x=angle BCD=angle PDC. So two angles equal the triangles are similar so angle QBP=x, and since angle QBA=x as well 2x=66degree.
Very quick stuff on this...Connect A with D. Now Angle X will be half of 66, as ADC arc is a circle with point B as center, AD is a chord of the circle. Hope this is useful.
Taking B as radius we can draw circle going through points A, D, C. So Angle ACD= 1/2 central angleABD by circles thm that any circumferential angle is half measure of central angle .So is 1/2(66)= 33 degrees.
We can use circle, wihf center in point B. 66 is a center angle, and x = 66/2. With together AD
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A,D,C on the circle(B,BA) Inscribed angle X(Angle ACD)equals half of central angle ABD:66/2=33.
I figured out 35° in 5 seconds according to what is given, so this is good enough for me.
A simple way is assume A, B, C are in one line, that means P and B are the same point. Due to BC = BD, angle BCD = angle BDC = 66/2 = 33.
The chord AD subtend the arc ADC. B is the center. So, the angle of ABD is the central angle of the chord AD. and the angle of ACD(x) is angle of the circumference of the chord A D. X =66÷2=33
thank you lot's
Since BA=BD=BC, one can draw a circle centered at B, with radius BA=BD=BC. Then, by inspection, since the 66 degree angle, ABD, is the central angle corresponding to the arc AD, it is twice the angle ACD, corresponding to the same arc. Hence the angle ACD is one half of angle ABD,i.e., 33 degrees.
I've marked ∠BAC as b, ∠DBC as c.
2b+c+66=180 → c=114-2b
2(b+x)+c=180 → c=180-2x-2b
114-2b=180-2x-2b → x=33
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@@PreMath Thank you! You are awesome too!
180 - 66 = 114
@@ludmillamerdian9089
Thank you! typo... fixed!
Point B is center of circle and points C, D, A are on this circle... Angle ABD is twice larger than angle ACD. ;)
B is center of circle.
A, D, C is point on the circumference.
angle ABD is 66°.
so, angle ACD is 33°.
*Q.E.D.*
i did this by myself...happy..
Draw circle with centre B radius AD. Therefore X is half the angle that cord AD subtends at centre. X=66/2=33
The inscribed angle theorem states that an angle θ inscribed in a circle is half of the central angle 2θ that subtends the same arc on the circle & central angle ABD = 66° therefore angle ACD (X) = 33° _ both are subtending the same arc AD
Another way _ we have angle APB = angle CPD & triangle sum = 180°
Therefore BÂC + 66 = X + angle BDC & (angle BDC = X + BÂC) ==> 66 = 2X ==> X = 33
How to calculate the value of Alpha in this ?
In MVC’s solution x = angle ACD, not BCD, intercepts the chord AD. But the conclusion is correct.
Another way
Here BC, BD and BA R equal
So we can draw a circle passing through point A, D, C. Hence angle ACD= half angle ABD.= 33°{Central angle is twice the angle on remaining part}.
If AC and BD are both segments of straight lines and this is a plane, how is this possible? AB, BP, and BC are all the same length, which should make B the center of a circle with AB, BP, and BC as radii, right?
B is center of circle circumferencing A D and C so x is half of 66 … relation of central angle and circumference angle at same arc… so x is equal to 33….
Я немного по другому решал. Дольше, чем Вы, но ответ такой же. Я просто пересчитывал все углы. Спасибо за краткость)
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Спасибо, что поделились! Ваше здоровье!
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Can you do a video about differentiation dy/dx
circumference C (B) center B, radius AB = BD = BC
AD = chord = arc circumference C (B)
Angle x with vertex C on circumference C (B) insists on arc AD, with respective angle in the center = 66 ° = 2x
Hence x = 33 °
АВ, ВD и ВС - радиусы описанной окружности с центром в точке В. Дуга АВD =66 градусов. Дуга угла АСD в два раза меньше или 33.
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I had convinced myself that if the angle DBC was less or greater then the angle x would change , so I got out my trusty ruler and protractor and carefully drew a number of different diagrams. x was always 33. Sorry to have doubted you ! ! ! !
Very nice question and explanation thnx
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another solution the the 3 point ADC AT perimeter of circle which its center @ B , CENTERAL ANGLE 66 = TWICE circumferential angle x HENCE x=33
Super very nice explanation thank you sir
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Hi sir i am your big fan as millions.
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To my mind it is better to replace the value 66° with a parameter from the interval ]0, π[.
66+M+2N = 180
M+2(X+N) = 180 and M+2N = 180-2X
66+(180-2X) = 180
X = 33°
The main key is recognizing the two isosceles triangles.
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That was my method as well.
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I remeber this in school.
Assume the angle next to x is y, and the angle next to 66 one is z, as the sum of interior angles of any triangle is 180, 2(x+y) + z = 2y + z + 66 , x= 33
Since AB=BD=BC, then A,D and C are cyclic and B is the center of the circle. Then x= Angle DCP = 1/2 of the angle ABD=1/2. 66=33.
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How to draw the quadrilateral ABCD? What is the angle of ABC?
les points aA C et D sont sur un cercle de centre B
soit E l’extrémité du diamètre DBE, les angles ACD =x et AED sont égaux
donc l'angle AED =ABD/2 =66°/2=33° et don
c x=33°
Super travail! Excellente approche
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B is the center of the circle
make a circle
A,D,C On the circumference
.....
You are not drawing proportionally. If BA = BP = BC, then APC can not be a straight line, or can it?
66÷2＝33
A,C,D, point is in same cycle, B is cycle center.
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Super duper bumper easy question!
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Inscribed angles 66 / 2 = 33
That's how I solved it, too! 😉
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Hmmmm! This one is a little tougher. Ah!!! I found angle X - It is right on the Drawing in RED. Next question.
I solved it
Dude
I thought segment BP is congruent to AB and BC😭
I was a bit confused when using isosceles triangle theorem, and why it does not match up
Using circles theorom x= 66/2 = 33 degrees.
يمكن رسم دائرة مركزها b تمر من النقاطaوcوdيكون قياسxالمحيطية =نصف قياسالمركزية66المشتركةمعها بالقوس و=٣٣ اعلم انكم تعلمون بهذه الطريقة ولكني احببت ان اشارك وشكرا لكم
Es más fácil usando ángulo inscrito.
En Perú es hasta un teorema, le dicen
"TEOREMA DE LAS TRES PIERNAS"
Just draw a circle centered b then 66/2=33
CLEVER ...😏😏
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2x = 66: x = 33
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What if ∠ABD=179°59'59.99...9''?
How do uou know that ab=bc=bd???
I was off by 3 degrees I just did 45-15 and got 30, so it's 48- 15 = 33
AB=BD=BC , raggi uguali di una stessa circonferenza con centro in B, angolo X = angolo alla circonferenza dell'arco AD. Angolo ABD = 66° quale angolo al centro dello stesso arco AD quindi X = 1/2 di 66 = 33°
Since BD, BC, BA are equal....
Points A, D , C are in the circumference with center B.
Central angle ABD determines an arc AD = 66. Then angle ACD is half the arc AD... Done.
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AD/sin66 = BD/sin(24+Alpha) = AB/sin(90-Alpha); AB = BD => 24+Alpha=90-Alpha => Alpha =33
Окружность и вписанный угол- половина центрального.
33 degree
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以B点为圆心，BC为半径画圆，×=∠ABD/2=33°
A quoi sert le theoreme de l'angle inscrit? Qui est la moitié de l'1ngle au centre . Les trois points A,C et D sont sur le même cer le de centre . L'arc AD de ce cercle est intercepté par l'angle ACD qui est la moitié de l'angle au centre ABD interceptant le meme arc AD du cercle....
INSCRIBED ANGLE.