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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