In an isosceles triangle abc with base ac equal to 10, the cosine of angle A is 0.8 find the area of the triangle.

Given:

ABC – isosceles triangle;

AB = BC;

AC is the base.

AC = 10;

cos A = 0.8;

Find the area of ​​the triangle.

Solution:

1) In order to find the area of ​​the triangle ABC, we use the formula:

S = ½ * AC * BH, where BH is the height drawn from point B to the base of the AC.

Height divides an isosceles triangle into 2 right-angled triangles.

2) Consider a triangle ABN, angle H – 90 degrees, AH = ½ * AC = ½ * 10 = 5.

sin a = √ (1 – 0.8 ^ 2) = √ (1 – 0.64) = √0.36 = 0.6;

cos a = AH / AB, hence AB = AH / cos a = 5 / 0.8 = 5 / (4/5) = 5/1 * 5/4 = 25/4;

sin a = BH / AB, hence BH = AB * sin a = 25/4 * 0.6 = 25/4 * 6/10 = 25/4 * 3/5 = 5 * ¾ = 15/4;

3) Find the area S.

S = ½ * AC * BH = ½ * 10 * 15/4 = 5 * 15/4 = 75/4 = 18.75.

Answer: S = 18.75.