Question 91

In triangle ABC, ∠ABC = $$15^o$$ .D is a point on BC such that AD = BD. What is the measure of ∠ADC (in degrees)?

Solution

Given : ∠ABC = $$15^o$$ and AD = BD

To find : ∠ADC = $$\theta$$ = ?

Solution : AD = BD, => $$\angle$$ ABD = $$\angle$$ DAB = $$15^o$$

=> $$\angle$$ ABD + $$\angle$$ DAB = $$15+15=30^\circ$$

Using, exterior angle property of a triangle,

=> $$\angle$$ ADC = $$\angle$$ ABD + $$\angle$$ DAB

=> $$\theta=30^\circ$$

=> Ans - (B)

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