Question

I need help with Algebra 2!

Answer 1

For sin, cos, and tan, let's use an abbreviation.

SOH-CAH-TOA

`sin=(opposite)/(hypotenuse)`

`cos=(adjacent)/(hypotenuse)`

`tan=(opposite)/(adjacent)`

But before that, we should solve for the missing side.

Since this triangle is a right triangle, we can use the Pythagorean Theorem.

`a^2+b^2=c^2 \\ \\ 6^2+8^2=c^2 \\ \\ c^2=36+64 \\ \\ c^2=100 \\ \\ c=10`

So the hypotenuse is 10.

Let's plug in.

sin θ = 6/10 --> 3/5

cos θ = 8/10 --> 4/5

tan θ = 6/8 --> 3/4

Now for the cosectant, secant, and cotangent, we'll just use the definitions.

Cosectant is hypotenuse over opposite.

Secant is hypotenuse over adjacent.

Cotangent is adjacent over opposite.

So they are basically the inverses.

csc θ = 10/6 --> 5/3

sec θ = 10/8 --> 5/4

cot θ = 8/6 --> 4/3