Find the Value of Angle X in this Triangle | Fast & Easy Tutorial

[nickeleytheriou]

a pure geometrical solution:
draw the bisector AE of angle CAD (E on BC).
The triangle AEB is isosceles so AE = EB
The triangles DEB and CAE are equal ( BD=AC, BE=AE, angle DBE = angle CAE), so: angle EDB = angle ACE
which means that the quadrilateral ACED is writable in a circle.
The inscribed angles DAE and DCE target to same arc DE, so:
angle DAE = angle DCE = 20°

[philipkudrna]

Very creative and finally something other than pythagoras and quadratic equations! Thank you!

[harikatragadda]

Another way to solve it.
Drop a line CE between A and D, such that angle CEA = 40°, therefore CA= CE
By Exterior Angle Theorem, CEA= angle EBC + angle ECB
Therefore angle ECB= 20°
∆CEB is Isosceles and CE=EB
Since CE = CA= DB
We have EB=DB (E coincides with D)
Therefore, X=20°

[fatjonnocka]

Construct DE //AC and
DE = AC.Angle CAD = Angle ADE = 40° like interior alternative angles.Since ED = DB then triangle EDB is iscoceles and anggle DEB = 40/2 = 20°.Quadrileteral BCDE is a balloon so x = angle DEB = 20°

[rohankumarroy]

Sir, If we draw bisector of ang.(40°) and complete the triangle, we will get to congruent triangle and after that question get easily solved using ASP

[tomcruise]

It was one of the best questions.
In India, there is an exam namely CAT for the Management of Business Administration (MBA). A similar question was asked in CAT 2003-04, to the best of my memory.

[haroldwood]

Beautifully clear work. Thank you. It was a pleasure to watch.

[kalumindika]

as a student your help is essential thing for my education. thanks sir. please help as well in this manner

[JLvatron]

Great video! I had figured that side CD was the same length as AC, so ACD was Isoceles, and this made BCD also Isoceles with x=20. All the angles made sense, but I couldn't prove that they HAD to be.
Great method. In the words of the late Owen Hart, "I loved it!"

[kartikeyanchelladurai]

Thank you very much PREMATH SIR 🙏🙏👍👍If I'm weak in any topic of Maths...I will ask you in comment 😊So, I hope you will send videos about that topic🙏🙏👍👍😃😃😁😁

[michaelkouzmin]

I solved the problem applying the Law of Sines to triangles ACD and BCD: AC/CD = sin(ADC)/sin(40); BD/CD=sin(x)/sin(20); ADC = 20+x; so sin(x)/sin(20) = sin(20+x)/sin(40).
What gives us x=20.

[simons]

What does mean congruent? In translator it said same(in my language), but they dont look the same.

[vladimirpoleshchuk]

Можно отразить треугольник ВСД относительно СВ. Получаем равнобедренную трапецию. Угол Х будет равен 20 как накрест лежащий. Это проще и быстрей

[SwainLake]

Using sine theorem and trig identities gets us to tan x = tan 20. So x= 20 degrees. Easy and quick!

[yes]

I solved same problem by another method. I used sin angle/side formula used in triangles.. I got same result...

[adeshchopra]

Wonderful problem! Hoping to see more of such gems!

[flbeblue]

I made an educated guess and for once, mathematically, it was correct! Helps that I wasn’t born or educated in Florida where facts science math and books are becoming n/a ty DeSatan.

[hectormoloko]

The beauty about mathematics is that there are many ways to skin a cat. Great problem to solve, I used the sin and cosine rules and the fact that AC = DB and I got to the same solution. After all LOGIC must prevail in all that we do :).

[dhrubajyotidaityari]

Equal sides l, common side m,
We have from two triangles,
l/{sin(X+20}=m/sin40, and
l/{sin X}=m/sin20
Solving, X=20 Ans.

[kevinmadden]

Use the Law of Sines in triangle
ADC and triangle BDC and utilized the the fact that AC equals BD and angle ADC equals (x plus 20).