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
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