In a triangle ABC, the angle is C = 90. tgA = √15. find sinB.

Let us express the tangent of angle A: tgA = BC / AC = √15.

That is, the BC side refers to the AC side as √15: 1. Let AC = x, then BC = x√15.

By the Pythagorean theorem: AB² = AC² + BC² = х² + (х√15) ² = х² + 15х² = 16х².

Hence, AB = √ (16x²) = 4x.

Let us express the sine of angle B: sinB = AC / AB.

Since AC = x and AB = 4x, then sinB = x / (4x) = 1/4 = 0.25.

Answer: sinB = 0.25.



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