How do you find the five remaining trigonometric function satisfying sintheta=3/8sinθ=38, costheta<0cosθ<0?

1 Answer
Jun 4, 2017

cosΘ=(-√55)/8
tanΘ=(-3√55)/55
cscΘ=8/3
secΘ=(-8√55)/55
cotΘ=(-√55)/3

Explanation:

Since sin is positive and cos is less than 0, it means that this triangle is in quadrant 2.

Use Pythagorean's theorem (a^2+b^2=c^2) and plug in your given values: a=3 and c=8 and solve.

3^2+b^2=8^29+b^2=64b^2=55b=-√55 (it is negative because it is in quadrant 2 and the cos is the x value of the triangle)

Now that you have all side lengths, simply plug them in to the remaining trigonometric functions:

cosΘ=(-√55)/8
tanΘ=3/(-√55)=(-3√55)/55
cscΘ=8/3
secΘ=8/(-√55)=(-8√55)/55
cotΘ=(-√55)/3