Finding the other ratios from one

One ratio is enough

Suppose you know only that $\sin A = \dfrac{5}{13}$. That tells you the shape of the triangle, though not its size.

1. Draw the sides. Sine is opposite over hypotenuse, so take the opposite side as $5k$ and the hypotenuse as $13k$. Here $k$ is any positive number.
2. Find the third side. By Pythagoras, adjacent $= \sqrt{(13k)^2 - (5k)^2} = \sqrt{144k^2} = 12k$.
3. Write the ratios. The $k$ cancels in every fraction.

Many students skip the $k$ and just use $5$ and $13$. That works too, because the ratios do not depend on size. The $k$ reminds you why.

A worked example with a reciprocal

Given $4 \sec \theta = 5$, find $\sin \theta$ and $\tan \theta$.

1. Rewrite. $\sec \theta = \dfrac{5}{4}$, so $\cos \theta = \dfrac{4}{5}$.
2. Draw. Adjacent $= 4k$, hypotenuse $= 5k$.
3. Pythagoras. Opposite $= \sqrt{25k^2 - 16k^2} = 3k$.
4. Answer. $\sin \theta = \dfrac{3}{5}$ and $\tan \theta = \dfrac{3}{4}$.

Working out an expression

Once you have every ratio, you can work out any expression in them.

A worked example. If $\tan A = \dfrac{7}{24}$, find $\sin A + \cos A$.

1. Draw. Opposite $= 7k$, adjacent $= 24k$.
2. Hypotenuse. $\sqrt{49k^2 + 576k^2} = 25k$.
3. Ratios. $\sin A = \dfrac{7}{25}$ and $\cos A = \dfrac{24}{25}$.
4. Add. $\sin A + \cos A = \dfrac{31}{25}$.

Ratios in everyday life

A ladder. A ladder $6.5\ \text{m}$ long leans against a wall, with its foot $2.5\ \text{m}$ from the wall. Let $\theta$ be its angle with the ground.

1. The height. $\sqrt{6.5^2 - 2.5^2} = \sqrt{42.25 - 6.25} = \sqrt{36} = 6\ \text{m}$.
2. The ratios. $\cos \theta = \dfrac{2.5}{6.5} = \dfrac{5}{13}$, $\sin \theta = \dfrac{6}{6.5} = \dfrac{12}{13}$ and $\tan \theta = \dfrac{6}{2.5} = \dfrac{12}{5}$.

A ramp. A wheelchair ramp is $4\ \text{m}$ long and rises $0.6\ \text{m}$. The ramp is the hypotenuse and the rise is opposite the slope angle, so its sine is $\dfrac{0.6}{4} = 0.15$. A smaller sine means a gentler ramp.

Practise with new numbers

Try it in Python

This program starts from $\sin A = \dfrac{5}{13}$, finds the third side and prints the other five ratios.

Common mistakes

• Using the given numbers for the wrong sides. In $\cos A = \dfrac{4}{5}$, the $4$ is the adjacent side, not the opposite one.
• Adding squares when you should subtract. When the hypotenuse is known, subtract to find a shorter side.
• Forgetting the reciprocal step. For $\sec \theta = \dfrac{5}{4}$, turn it over to get $\cos \theta$ first.
• Expecting whole numbers. A side can be a root such as $\sqrt{5}$; the ratios still work.

Remember these

• One ratio fixes the shape: draw its two sides as multiples of $k$.
• Pythagoras gives the third side; then all six ratios follow.
• The $k$ cancels, so the size of the triangle never matters.
• In a real problem, decide which length is the hypotenuse, the opposite side and the adjacent side.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. If $\cos A = \dfrac{3}{5}$, what is $\sin A$?

2. If $\cot B = \dfrac{12}{5}$, what is $\operatorname{cosec} B$?

3. If $\sin A = \dfrac{1}{3}$, what is the adjacent side when the hypotenuse is $3$?

4. A slide is $5\ \text{m}$ long and its top is $3\ \text{m}$ above the ground. What is the sine of its angle with the ground?

5. Why can you take the sides as $5$ and $13$ instead of $5k$ and $13k$?