In the given figure find the value of x ?
Thank you!

EXPLANATION:

In ∆ ABC, ∠CAB = 65°

AB = AC

⇛∠ACB =∠CBA

Since the angles opposite to equal sides are equal.

Let ∠ACB = ∠CBA = A°

We know that

The sum of all angles in a triangle is 180°

⇛∠ACB + ∠CBA + ∠CAB = 180°

⇛A°+A°+65° = 180°

⇛2A°+65° = 180°

⇛2A° = 180°-65°

⇛2A° = 115°

⇛A° = 115°:2 =115°/2

⇛A° = 57.5°

∠ACB =∠CBA =57.5°

and

In ∆ BCD , ∠BCD = 110°

and

BC = CD

⇛∠BDC =∠CBD

Since the angles opposite to equal sides are equal.

Let ∠BDC =∠CBD = M°

The sum of all angles in a triangle is 180°

∠BCD+∠BDC + ∠CBD = 180°

⇛M°+M°+110° = 180°

⇛2M°+110°° = 180°

⇛2M° = 180°-110°

⇛2M° = 70°

⇛M° = 70° :2 =70°/2

⇛M° = 35°

.°. ∠BDC = ∠CBD = 35°

Now,

∠B = ∠ABC+ ∠CBD

⇛x = 57.5°+35°

⇛x = 92.5°

.°. x = 92.5°

Answer: The value of x is 92.5°.

Additional comment: •The sum of all angles in a triangle is 180°

• the angles opposite to equal sides are equal. •The exterior angle of a triangle formed by extending one side is equal to the sum of the opposite interior angles.

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