In ∆ABC, ∠Α = 50°, ∠B = 70° and bisector of ∠C meets AB in D (Fig. 6.17). Measure of ∠ADC is
(a) 50°
(b) 100°
(c) 30°
(d) 70°

Solution:

Given, ABC is a triangle.

∠Α = 50° and ∠B = 70°

The bisector of ∠C meets AB in D.

We have to find the measure of ∠ADC.

Considering triangle ADC,

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, ∠ADC + ∠DAC + ∠DCA = 180°

∠ADC + 50° + ∠DCA = 180°

∠ADC + ∠DCA = 180° - 50°

∠ADC + ∠DCA = 130°

∠ADC = 130° - ∠DCA ---------------------- (1)

Considering triangle BDC,

The exterior angle theorem states that the measure of an exterior angle is equal to the sum of the measures of the two opposite interior angles of the triangle.

By exterior angle property,

∠ADC = ∠DBC + ∠BCD

∠ADC = 70° + ∠BCD

We know ∠BCD = ∠ACD

So, ∠ADC = 70° + ∠ACD

From (1),

∠ADC = 70° + 130° - ∠ADC

∠ADC + ∠ADC = 200°

2∠ADC = 200°

∠ADC = 200°/2

Therefore, ∠ADC = 100°

✦ Try This: In a triangle ABC, ∠A + ∠B = 144° and ∠A + ∠C = 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 41

In ∆ABC, ∠Α = 50°, ∠B = 70° and bisector of ∠C meets AB in D (Fig. 6.17). Measure of ∠ADC is: (a) 50°, (b) 100°, (c) 30°, (d) 70°

Summary:

In ∆ABC, ∠Α = 50°, ∠B = 70° and bisector of ∠C meets AB in D (Fig. 6.17). The measure of ∠ADC is 100°.

---

☛ Related Questions:

• 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) . . . .
• If D is the midpoint of the side BC in ∆ABC where AB = AC, then ∠ADC is: (a) 60°, (b) 45°, (c) 120°, . . . .