In a right-angled triangle ABC with a right angle, C sinA = 0.8. Find sinB.

First way.

Since triangle ABC is rectangular, the sine of one of its acute angles is equal to the cosine of its other acute angle.

sinA = cosB = 0.8.

Then Cos2B + Sin2B = 1.

Sin2B = 1 – Cos2B = 1 – 0.64 = 0.36.

SinB = 0.6.

Second way.

The sine of an acute angle of a right triangle is the ratio of the opposite leg to the hypotenuse.

Let BC = 8 cm, AB = 10 cm, then sinA = 8/10 = 0.8.

By the Pythagorean theorem, AC ^ 2 = AB ^ 2 – BC ^ 2 = 100 – 64 = 36.

AC = 6 cm.

Then SinB = AC / AC = 6/10 = 0.6.

Answer: SinB = 0.6.