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.

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