Question

In triangle ABC, ∠ABC=70° and ∠ACB=50°. Points M and N lie on sides AB and AC respectively such that ∠MCB=40° and ∠NBC=50°. Find m∠NMC.

Answer 1

Answer:

∠NMC  = 50°

Step-by-step explanation:

The interpretation of the information given in the question can be seen in the attached images below.

In ΔABC;

∠ A + ∠ B + ∠ C = 180°    (sum of angles in a triangle)

∠ A + 70°  + 50°  = 180°

∠ A = 180° - 70° - 50°

∠ A =  180° - 120°

∠ A =  60°

In ΔAMN ; the base angle are equal , let the base angles be x and y

So; x = y   (base angle of an equilateral  triangle)

Then;

x + x + 60° = 180°

2x +  60° = 180°

2x = 180° - 60°

2x = 120°

x = 120°/2

x = 60°

∴ x = 60° , y = 60°

In ΔBQC

∠a + ∠e + ∠b = 180°

50° + ∠e + 40° = 180°

∠e = 180° - 50° - 40°

∠e = 180° - 90°

∠e = 90°

At point Q , ∠e = ∠f = ∠g = ∠h = 90°  (angles at a point)

∠i  = 50° - 40° = 10°

In ΔNQC

∠f + ∠i   + ∠j = 180°

90° + 10° + ∠j = 180°

∠j  = 180° - 90°-10°

∠j  = 180° - 100°

∠j  = 80°

From  line AC , at point N , ∠y + ∠c + ∠j = 180°   (sum of angles on a straight line)

60° + ∠c + ∠80° = 180°

∠c  = 180° - 60°-80°

∠c  = 180° - 140°

∠c  = 40°

Recall that :

At point Q , ∠e = ∠f = ∠g = ∠h = 90°  (angles at a point)

Then In Δ NMC ;

∠d + ∠h + ∠c = 180°   (sum of angles in a triangle)

∠d + 90° + 40° = 180°

∠d  = 180° - 90° -40°

∠d  = 180° - 130°

∠d  = 50°

Therefore, ∠NMC = ∠d  = 50°