In the triangle ABC AC = BC = 10, cosA = 0.6. Find the area of the triangle.

In the ABC triangle it is known:

Sides AC = BC = 10;
cos A = 0.6.
Find the area of the triangle.

Decision:

1) The ABC triangle is isosceles.

CM is the height of an isosceles triangle.

2) Consider a triangle AFM with a right angle M.

sin a = CM / AC;

CM = AC * sin a = 10 * √ (1 – sin ^ 2 a) = 10 * √ (1 – 0.6 ^ 2) = 10 * √ (1 – 0.36) = 10 * √0.64 = 10 * 0.8 = 8;

3) cos a = AM / AC;

AM = AC * cos a = 10 * 0.6 = 6;

4) AB of triangle ABC is equal to:

AB = 2 * AM = 2 * 6 = 12.

5) Find the area of the rectangle:

S = 1/2 * CM * AB = 1/2 * 8 * 12 = 8/2 * 12 = 4 * 12 = 48.

Answer: S = 48.