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x= Angle A= Sin-1(3/5)= 36.9 degrees.

But, by manipulating angles, we get 50 degrees.

There is an error in the question.

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vb1991
This is a very poor quality question and the answer is wrong.

Since the larger triangle is a 3-4-5 type Pythagorean triangle with angle B = 90°, then the following HAS to be true:

angle A = 30°
angle C = 60°

vb1991

Mate, not all right angle triangles have measure of angles : 30, 60 and 90.

All right angled triangle will follow Pythagoras Theorem. But for 30-60-90 type right triangle sides will be in ratio : 1 : \sqrt{3} : 2

Clearly, 3-4-5 do not follow above side ratio.

So, angles will not be in 30-60-90 form for 3-4-5 right triangle.

In fact for 3-4-5 right triangle corresponding angles are 36.87 - 53.13 - 90.

Hope this helps!!
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vb1991
This is a very poor quality question and the answer is wrong.

Since the larger triangle is a 3-4-5 type Pythagorean triangle with angle B = 90°, then the following HAS to be true:

angle A = 30°
angle C = 60°

vb1991

Mate, not all right angle triangles have measure of angles : 30, 60 and 90.

All right angled triangle will follow Pythagoras Theorem. But for 30-60-90 type right triangle sides will be in ratio : 1 : \sqrt{3} : 2

Clearly, 3-4-5 do not follow above side ratio.

So, angles will not be in 30-60-90 form for 3-4-5 right triangle.

In fact for 3-4-5 right triangle corresponding angles are 36.87 - 53.13 - 90.

Hope this helps!!

I conflated two separate concepts, my bad. Thanks for the explanation!

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Please explain how is ∠CDB = ∠BDC = 65°?
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Architmahajan25
Please explain how is ∠CDB = ∠BDC = 65°?

It's already given in the problem question that BDC is approximately 65°. Since CDB and BDC are even, CDB is also 65.
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