Question

In Fig. 4.58, ∆ABC is a triangle such that AB/AC = BD/DC, ∠B=70°, ∠C=50°. Find the ∠BAD.

Answer 1

SOLUTION :  

Given : in Δ ABC, AB/AC = BD/DC, ∠B = 70° and ∠C = 50°

In Δ ABC,

∠A + ∠B +∠C = 180°

∠A + 70° + 50° = 180°

∠A = 180 - (70° + 50°)

∠A = 180° - 120°

∠A = 60°

Since, AB/AC = BD/DC

Therefore, AD is the bisector of ∠A

[If a line through one vertex of a triangle divides the opposite sides in the ratio of other two sides, then the line bisects the angle at the vertex.]

∠BAD = 60°/2 = 30°

Hence, ∠BAD = 30°.

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Answer 2

Answer:

We have, if a line through one vertex of a triangle divides the opposite side in the ratio of the other two sides, then the line bisects the angle at the vertex. ∴ ∠1 = ∠2 In ΔABC ∠A + ∠B + ∠C = 180° ⇒ ∠A + 70 ° + 50° = 180° [∵ ∠B = 70° and ∠C = 50°] ⇒ ∠A = 180° − 120° = 60° ⇒ ∠1 + ∠2 = 60° ⇒ ∠1 + ∠1 = 60°    [∵ ∠1 = ∠2] ⇒ 2∠1 = 60° ⇒ ∠1 = 30° ∴ ∠BAD = 30

Step-by-step explanation:

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