In triangle ABC, angle C is 90. AB is 10. sinA is 4/5. Find the AC.

Given: right-angled triangle ABC;

angle C = 90;

sin A = 4/5;

AB = 10;

Find: AB -?

Decision:

1) Let’s use the basic trigonometric formula:

cos ^ 2A + sin ^ 2A = 1;

cos ^ 2A = 1 – sin ^ 2A;

cos ^ 2A = 1 – 16/25 (represent 1 as an ordinary fraction with a denominator of 25);

cos ^ 2A = 25/25 – 16/25;

cos ^ 2A = 9/25;

cos A = 3/5;

2) Consider a right-angled triangle ABC. The cosine of the angle in a right-angled triangle is equal to the ratio of the adjacent leg to the hypotenuse, that is, equal to the ratio of AC to AB. Hence:

cos A = AC / AB;

AC = cos A * AB;

AC = 10 * 3/5;

AC = (10 * 3) / 5 (reducing);

AC = (2 * 3) / 1;

AC = 6

Answer: AC = 6.