How would I find sin B, Cos B, sin C, and Cos C

Question

asked Dec 12, 2023 210k views

1 Answer

Loicgasser · Answer 1 · 2023-12-18T05:34:30+0000

The given triangle is

We get hypotenuse =BC=40 by observing the given triangle.

Consider the angle B.

For angle B, the Opposite side is AC=32 adjacent side AB =24.

We know that

$\sin \theta=\frac{Opposite\text{ side}}{\text{Hyponetuse}}$

$\sin B=(AC)/(BC)$

Substitute AC=32 and BC=40, we get

$\sin B=(32)/(40)=0.8$

We know that

$\cos \theta=\frac{adjacent\text{ side}}{\text{Hypotenuse}}$

$\cos B=(AB)/(BC)$

Substitute AB=24 and BC=40, we get

$\cos B=(24)/(40)=0.6$

Consider the angle C.

For angle C, the Opposite side is AB=24 and the adjacent side AC=32.

$\sin C=(AB)/(BC)$

Substitute AB=24 and BC=40, we get

$\sin C=(24)/(40)=0.6$

$\cos C=(AC)/(BC)$

Substitute AC=32 and BC=40, we get

$\cos C=(32)/(40)=0.8$

Hence the answers are

$\sin B=0.8$

$\cos B=0.6$

$\sin C=0.6$

$\cos C=0.8$