One ratio gives all the others

Know one ratio and you can find all six. Draw the right triangle and fill in the third side with Pythagoras.

Example: sin A = 1/3. sin is opposite/hypotenuse, so call the opposite side k and the hypotenuse 3k. Use k because a ratio only fixes the proportion, not the actual lengths.

• The third side is √(9k² – k²) = √(8k²) = 2√2 k. A length can't be negative, so take the positive root.
• Now read off the rest: cos A = 2√2/3, tan A = 1/(2√2), cosec A = 3, sec A = 3/(2√2), cot A = 2√2.

Example: tan A = 4/3. The sides are 4k and 3k, so the hypotenuse is 5k. Then sin A = 4/5 and cos A = 3/5.

sin A and sin B in the same triangle use different sides. Each angle has its own opposite side, so work it out separately for each.

What values are possible? The hypotenuse is the longest side, so:

So sin θ = 4/3 is impossible, but sec A = 12/5 is fine.

Equal sines mean equal angles. If two acute angles A and B have sin A = sin B, then A = B. The same is true for cosines.