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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