Given a right-angled triangle ABC, angle C is a straight line. Find the sine of angle A if BC = 3, and the height lowered by the hypotenuse is 2.4.
Let’s denote the height dropped by the hypotenuse in the triangle ABC by MC.
Consider triangle CMB. Because CM is the height, then triangle CMB is rectangular. Let’s apply the Pythagorean theorem:
MB ^ 2 = CB ^ 2 – CM ^ 2
MB = √ (CB ^ 2 – CM ^ 2) = √ (9 – 5.76) = 1.8
Consider triangles ABC and CMB. They are both rectangular and corner B is common. It follows that triangles are similar in two angles.
Angle MCB = CAB.
sin (CAB) = sin (MCB)
Find the sine of the angle MCB from the triangle CMB. The sine of the angle is the ratio of the opposite leg to the hypotenuse.
sin (MCB) = MB / CB = 1.8 / 3 = 0.6
sin (CAB) = 0.6
Answer: sin (CAB) = 0.6
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