Find the angles of triangle ABC if angle B is 40⁰ greater than angle A and angle C is five times greater than angle A.

Since the sum of all the angles of any triangle is 180º, the angle ∠B is 40º greater than the angle ∠A, and the angle ∠C is five times greater than the angle ∠A, then we express:

x is the degree measure of the angle ∠A;

x + 40 – degree measure of angle ∠В;

5x – degree measure ∠С;

x + 5x + x + 40 = 180;

x + 5x + x = 180 – 40;

7x = 140;

x = 140/7 = 20;

∠А = 20º;

∠В = 20º + 40º = 60º;

∠С = 20º · 5 = 100º.

Answer: angle А is equal to 20º, angle ∠В is equal to 60º, ∠С is equal to 100º.