In ∆ABC, ∠Α = 100°, AD bisects ∠A and AD⊥BC. Then, ∠B is equal to
(a) 80°
(b) 20°
(c) 40°
(d) 30°

Solution:

Given, ABC is a triangle.

AD bisects ∠A and AD⊥BC.

Also, ∠Α = 100°

We have to find the value of ∠B.

Since AD bisects ∠A,

∠DAB = ∠CAD = 50°

∠BDA = CDA = 90°

Considering triangle ABD,

By angle sum property of a triangle,

We know that the sum of all the three interior angles of the triangle is equal to 180 degrees.

So, ∠ABD + ∠DAB + ∠BDA = 180°

∠ABD + 50° + 90° = 180°

∠ABD + 140° = 180°

∠ABD = 180° - 140°

∠ABD = 40°

Therefore, ∠B is 40°.

✦ Try This: ∠P + ∠Q = 144° and ∠P - ∠Q = 124°. Calculate each angle of the triangle

☛ Also Check: NCERT Solutions for Class 7 Maths Chapter 6

NCERT Exemplar Class 7 Maths Chapter 6 Problem 40

In ∆ABC, ∠Α = 100°, AD bisects ∠A and AD⊥BC. Then, ∠B is equal to: (a) 80°, (b) 20°, (c) 40°, (d) 30°

Summary:

In ∆ABC, ∠Α = 100°, AD bisects ∠A and AD⊥BC. Then, ∠B is equal to 40°

☛ Related Questions:

• In ∆ABC, ∠Α = 50°, ∠B = 70° and bisector of ∠C meets AB in D (Fig. 6.17). Measure of ∠ADC is: (a) 50 . . . .
• If for ∆ABC and ∆DEF, the correspondence CAB ↔ EDF gives a congruence, then which of the following i . . . .
• In Fig. 6.18, M is the mid-point of both AC and BD. Then: (a) ∠1 = ∠2, (b) ∠1 = ∠4, (c) ∠2 = ∠4, (d) . . . .