Fine sin A, cos A, and tan A in the following triangle.

We know the hypotenuse and one leg, so let's use those to find the other leg first. Pythagoras, where you at?

a² + b² = c²

3² + b² = 6²

b² = 6² – 3²

b² = 36 – 9

b² = 27

Good deal. That's our side adjacent to angle A. The opposite side is 3, and the hypotenuse is 6. A dash of SOHCAHTOA should finish this problem off.

Find the six trig ratios of angle B.

Before we do any trig stuff, we're gonna need that hypotenuse. Let's fire up the Pythagorean Theorem.

a² + b² = c²

3² + 2² = c²

9 + 4 = c²

c² = 13 

Cool. That's our hypotenuse, and since we're looking at angle B, the opposite side is 2 and the adjacent side is 3. Time for a little SOHCAHTOA. Don't forget to rationalize those denominators.

For the reciprocal functions, we can just flip the first three over. Cosecant is the reciprocal of sine:

Secant flips cosine on its head:

And cotangent is the upside-down version of tangent.

Write cos(80°) in terms of its co-function.

Since sine and cosine are co-functions, we can just nab the sine of an angle complementary to A. Like so:

cos A = sin(90° – A)

cos(80°) = sin(90° – 80°)

cos(80°) = sin(10°)

Find x.

This guy is a 30°-60°-90° triangle. We know the long leg and we want the short leg, so all we've gotta do is divide by √3. And, of course, rationalize the denominator to make things pretty.

Find x.

We want x, which is the hypotenuse of a 30°-60°-90° triangle. The short leg is 15, so the hypotenuse is twice that.

x = 30

Find x.

Now we're dealing with a 45°-45°-90° triangle. We know the hypotenuse and we want one of those legs, so we divide by √2.

Find x if the unlabeled leg has a length of 4.

Let's not overthink it. The unlabeled leg and x are both legs of a 45°-45°-90° triangle. Since one leg is 4, then x = 4. These guys are always isosceles.