In an isosceles triangle ABC with base AC, the side AB is 10, and the height drawn

In an isosceles triangle ABC with base AC, the side AB is 10, and the height drawn to the base is 8. Find the cosine of angle A

Since CH is the height of the triangle ABC, the triangles AСН and BCH are rectangular.

In a right-angled triangle ABH, SinA = BH / AB = 8/10 = 0.8.

Then Cos2A + Sin2A = 1.

Cos2A = 1 – 0.82 = 1 – 0.64 = 0.36.

CosA = 0.6.

Answer: CosA = 0.6.



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