# Math sin/cos

posted by
**Anonymous**
.

On a piece of paper draw and label a right triangle using the given sides, solve for the unknown side and write the trigonometric functions for angles A and B, if a=5 and c=7.

I already found side b which equals 2 sqrts of 6.

Now I need to find the sin/cos of B

SinB

a)5/7

b)7/2sqrt6

c)2sqrt6/7

d)7/5

Cos B

a)7/5

b)2sqrt6

c)7/2sqrt6

d)5/7

If someone can tell me what values to put for sine and cosine I can do the rest. Thanks!

An easy way to remeber these rules is SOHCAHTOA. S is sine C is cosine and T is tangent. The other letters tell you what to use to find them. O is opposite A is adjacent and H is hypotoneuse. So the sine of B would be 2 sqrts of 6 (opposite) over 7(hypot.)

then to find cosine you would use 5(adjacent) over 7(hypot.)

would this still be the answer even though they were the same answers for angle A?

sin A = side opposite/hypotenuse = 5/7

sin B = side opposite/hypotenuse = 2(sqrt 6)/7

sin B and cos B will give the same answer for angle B. You can subtract angle B from 90o to obtain angle A or use sin A = a/c = 5/7

Angle A is angle A is angle A. =).

It matters little if you use sin, cos, tan, cot, or whatever, you will always get the same answers for angles A, B, and C as long as the sides don't change length.

I get the part you are saying but could you or someone help me with how to plug it in to a calculator and solve it? It'd help a bunch