In a triangle ABC, the angle is C = 90 degrees. SinA = √51 \ 10. Find SinB.

Given:

ABC – right triangle;

Angle C = 90 °;

sin A = √51 / 10;

Find sin B.

Decision:

1) First, find cos B.

Since cos B = BC / AB, then we get: sin A = BC / AB = cos B.

Hence, we obtain that cos B = sin A.

Substitute the known value sin A = √51 / 10 into the expression cos B = sin A and find cos B.

cos B = sin A = √51 / 10.

2) Find sin B by the Pythagorean theorem.

sin ^ 2 B = 1 – cos ^ 2 B = 1 – (√51 / 10) = 1 – 51/100 = √ (1 – 51) / √100 = √49 / √100 = √ (7/10) ^ 2;

From this we get sin B = 7/10.