# In triangle ABC AC = BC, AB = 10, the height AH is 3. Find the sine of the angle BAC.

Since in △ ABC the sides AC and BC are equal, this triangle is isosceles, then side AB is the base of an isosceles triangle, and ∠A and ∠B are the angles at the base of an isosceles triangle. Then:

∠A = ∠B.

Since ∠A and ∠B are equal, the sines of these angles will also be equal.

In △ AHB ∠AHB = 90 ° (since AH is the height), then the side AB, which lies opposite the right angle, is the hypotenuse of B AHB, and the sides AH and BH are cathets.

In a right-angled triangle, the sine of an acute angle is the ratio of the leg, which lies opposite this angle, to the hypotenuse. Opposite ∠B lies leg AH, then:

sin∠B = AH / AB.

By assumption, AH = 3, and AB = 10, then:

sin∠B = 3/10 = 0.3.

Since the sine of ∠B is equal to the sine of ∠A (aka ∠BAC), then:

sin∠A = 0, 3.

Answer: sin∠A = 0, 3.