The two sides of the triangle are 14 and 8 cm, and the cosine of the angle between them is 6/7. Find the area of this triangle.

From triangle ABM cosA = AM / AB.

6/7 = AM / 8;

AM = 6 * 8/7 = 48/7.

By the Pythagorean theorem AB ^ 2 = AM ^ 2 + BM ^ 2.

BM ^ 2 = AB ^ 2 – AM ^ 2 = 8 ^ 2 – (48/7) ^ 2 = (8 + 48/7) (8 – 48/7) = 8/7 * 104/7 = 832/49 …

BM = √ (832/49) = √ (64 * 13/49) = 8 * √13 / 7 = 8√13 / 7 (cm).

Find the area of the triangle:

S = ½ АС * ВМ = ½ * 14 * 8√13 / 7 = (14 * 8 * √13) / (2 * 7) = 8√13 (cm2).

2nd way.

Let’s apply the basic trigonometric identity.

sin2A = 1 – cos2A;

sinA = √ (1 – 36/49) = √ (13/49) = √13 / 7.

By the formula S = ½ AB * AC * sinA we find the area of the triangle.

S = ½ * 8 * 14 * √13 / 7 = 8√13 (cm2).

Answer: S = 8√13 cm2.