Q40 of 158 Page 159

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

Given:

In Δ ABC, ∠A = 100°, AD bisects ∠A and AD ⊥ BC

Formula Used/Theory:-

Pythagoras theorem:-

Base² + Height² = Hypotenuse²

When ∠A is bisected

Then;

∠DAB = 50° and ∠DAC = 50°

Then;

In Δ ABD

As AD ⊥ BC

∠ADB = 90°

Sum of all angles is 180°

∠ADB + ∠DAB + ∠B = 180°

90° + 50° + ∠B = 180°

∴ ∠B = 180° - 140° = 40°

Which of the following figures will have it’s altitude outside the triangle?

In Fig. 6.16, if AB ∥ CD, then

In ∆ABC, ∠A = 50°, ∠B = 70° and bisector of ∠C meets AB in D (Fig. 6.17). Measure of ∠ADC is.

If for ∆ABC and ∆DEF, the correspondence CAB ↔ EDF gives a congruence, then which of the following is not true?