In triangle ABC, angle C is 90 degrees, sinA = 11/14, AC = 10√3. find AB

In the triangle ABC, we find the hypotenuse AB, if it is known:

Angle C = 90 °;
sin A = 11/14;
AC = 10√3.
Decision:

1) Find the cosine of angle A.

cos A = √ (1 – sin ^ 2 a) = √ (1 – (11/14) ^ 2) = √ (1 – 121/196) = √ (196 – 121) / √196 = √75 / √196 = √3 * √25 / √196 = 5/14 * √3;

2) Knowing the cosine of angle A and the adjacent leg, we find the hypotenuse of the triangle.

cos A = AC / AB;

AB = AC / cos A = 10√3 / (5√3 / 14) = 10√3 * 14 / (5√3) = 10 * 14/5 = 10/5 * 14 = 2 * 14 = 28;

As a result, we got that the hypotenuse of the triangle is AB = 28.

Answer: AB = 28.