Solving for the Denominator
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- Solving for the Denominator - Video - Finding an unknown side (hypotenuse) Watch
Theory
Sometimes the side you want is on the bottom of the trig ratio, such as \(\sin 38^\circ = \dfrac{12}{x}\). This Year 9 maths page shows how to rearrange so that the last step is a division, how to tell which case you are in, and why the answer comes out bigger than the known side when it is the hypotenuse.
Label the sides and choose sin, cos or tan exactly as before. Then write the ratio and look where the unknown sits.
If the unknown is the denominator (on the bottom), multiply both sides by it, then divide by the ratio. The answer is the known side divided by the trig ratio.
This happens when you know a short side and want the hypotenuse, or when you know the opposite side and want the adjacent side with tan.
Sine and cosine of an acute angle are less than \(1\), so dividing by them makes the answer bigger. That fits, because the hypotenuse is the longest side.
Rearranging when the unknown is underneath:
The pattern:
| The ratio reads | Unknown is | Last step |
|---|---|---|
| \(\cos \theta = \dfrac{x}{20}\) | on top | \(x = 20 \cos \theta\) |
| \(\cos \theta = \dfrac{20}{x}\) | underneath | \(x = \dfrac{20}{\cos \theta}\) |
How to find a side when the unknown is underneath
- Label the sides from the given angle and choose the ratio.
- Write the ratio with the numbers in. If the unknown is on the bottom, this is the divide case.
- Rearrange: multiply both sides by the unknown, then divide by the ratio.
- Calculate in degree mode and round once, at the end.
- Check the hypotenuse is the longest side.
Opposite and hypotenuse, so use sin. This time \(x\) is underneath:
| \(\sin 38^\circ\) | \(=\) | \(\dfrac{12}{x}\) |
| \(x \sin 38^\circ\) | \(=\) | \(12\) |
| \(x\) | \(=\) | \(\dfrac{12}{\sin 38^\circ}\) |
| \(\) | \(=\) | \(19.491\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{19.5} \text{ cm}\) |
\(19.5 > 12\), as the hypotenuse must be.
Adjacent and hypotenuse, so use cos. \(18' = 0.3^\circ\).
| \(\cos 54.3^\circ\) | \(=\) | \(\dfrac{9}{y}\) |
| \(y\) | \(=\) | \(\dfrac{9}{\cos 54.3^\circ}\) |
| \(\) | \(=\) | \(15.423\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{15.4} \text{ m}\) |
Opposite and adjacent, so use tan. The unknown \(n\) is underneath:
| \(\tan 29^\circ\) | \(=\) | \(\dfrac{6.3}{n}\) |
| \(n\) | \(=\) | \(\dfrac{6.3}{\tan 29^\circ}\) |
| \(\) | \(=\) | \(11.365\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{11.4} \text{ cm}\) |
A small angle has a long adjacent side, so \(n > 6.3\) makes sense.
(i) the length of the cable \(c\) and
(ii) the distance \(g\) from the foot of the cable to the mast, each to one decimal place.
(i) The mast is opposite the \(47^\circ\) angle and the cable is the hypotenuse:
| \(\sin 47^\circ\) | \(=\) | \(\dfrac{21}{c}\) |
| \(c\) | \(=\) | \(\dfrac{21}{\sin 47^\circ}\) |
| \(\) | \(=\) | \(28.713\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{28.7} \text{ m}\) |
(ii) Use the given side again, not the rounded cable. Opposite and adjacent, so tan:
| \(\tan 47^\circ\) | \(=\) | \(\dfrac{21}{g}\) |
| \(g\) | \(=\) | \(\dfrac{21}{\tan 47^\circ}\) |
| \(\) | \(=\) | \(19.582\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{19.6} \text{ m}\) |
Common pitfalls
Frequently asked questions
How do you find the hypotenuse using trigonometry?
Use sin if you know the opposite side, or cos if you know the adjacent side. The hypotenuse is on the bottom, so divide the known side by the ratio.
What do you do when x is on the bottom of a trig ratio?
Multiply both sides by \(x\), then divide both sides by the ratio. The result is \(x = \dfrac{\text{known side}}{\text{ratio}}\).
Why is the answer bigger when you divide by sin?
For an acute angle, sin is a number between \(0\) and \(1\), and dividing by a number less than \(1\) makes the result bigger.
How do I know if I should multiply or divide?
Write the ratio with the numbers in. If the unknown is on top, multiply. If it is on the bottom, divide.