In triangle ABC, angle C is 90 degrees, CH-height, BC = 5, cosine of angle A = 7/25. find BH.

Determine the sine of the angle BAC.

Sin2BAC + Cos2BAC = 1.

Sin2BAC = 1 – Cos2BAC = 1 – 49/625 = (625 – 49) / 625 = 576/625.

SinBAC = 24/25.

Let us determine the length of the hypotenuse AB.

SinBAC = BC / AB.

AB = BC / SinBAC = 5 / (24/25) = 125/24 cm.

Triangles ACB and СBN are similar in acute angle, then:

AB / BC = BC / BH.

BH = BC2 / AB = 25 / (125/24) = 600/125 = 4.8 cm.

Answer: The length of the BH segment is 4.8 cm.



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