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Year 9 Maths Core Trigonometry

Finding a Side (Unknown Denominator)

20 practice questions 2 video lessons Theory + worked examples

Sometimes the unknown side ends up on the bottom of the ratio, as in \(\cos 42^\circ = \frac{11}{h}\). This NSW Year 9 Core maths subtopic, under NESA outcome MA5‑TRG‑C‑01, deals with that case: the last step becomes a division, \(h = \frac{11}{\cos 42^\circ}\), instead of a multiplication.

You will recognise when the unknown is the denominator, rearrange by multiplying up and then dividing, and use the quick form of the known side divided by the ratio. You will round to one or two decimal places, work from statements that name the vertices, and see why dividing by a sine makes the answer larger. Problems include the length of a support cable, how far someone stands from a building, and two triangles that share a side. It is part of Year 9 Core Trigonometry.

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  • Finding a Side (Unknown Denominator) - Video - Finding Missing Sides Watch
  • Finding a Side (Unknown Denominator) - Video - Finding an unknown side 2 (hypotenuse) Watch
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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.

Unknown on top compared with unknown underneathTwo triangles with a 35 degree angle. In the first the hypotenuse is 12 and the opposite side x is wanted, so x equals 12 sin 35 degrees. In the second the opposite side is 12 and the hypotenuse x is wanted, so x equals 12 divided by sin 35 degrees. 35° 12 x x on top: x = 12 sin 35° 35° x 12 x underneath: x = 12 ÷ sin 35°
Same triangle shape, same ratio. Only the position of the unknown decides whether to multiply or to divide.
Dividing by sine makes the answer biggerA bar of length 12 for the known short side and a longer bar of about 20.9 for the hypotenuse, 12 divided by sin 35 degrees. 12 (known short side) 12 ÷ sin 35° ≈ 20.9 (hypotenuse) dividing by a number less than 1 makes it bigger
\(\sin 35^\circ \approx 0.57\) is less than \(1\), so dividing by it gives a bigger number, as the hypotenuse must be.

Rearranging when the unknown is underneath:

\[\sin \theta = \dfrac{k}{x} \quad \Rightarrow \quad x \sin \theta = k \quad \Rightarrow \quad x = \dfrac{k}{\sin \theta}\]
x=ksinθ

The pattern:

\[\text{unknown} = \dfrac{\text{known side}}{\text{the ratio}}\]
unknown=known sideratio
The ratio readsUnknown isLast step
\(\cos \theta = \dfrac{x}{20}\)on top\(x = 20 \cos \theta\)
\(\cos \theta = \dfrac{20}{x}\)underneath\(x = \dfrac{20}{\cos \theta}\)
Two triangles sharing a side. Find the shared side from the first triangle, keep it unrounded, then use it as the known side in the second.

How to find a side when the unknown is underneath

  1. Label the sides from the given angle and choose the ratio.
  2. Write the ratio with the numbers in. If the unknown is on the bottom, this is the divide case.
  3. Rearrange: multiply both sides by the unknown, then divide by the ratio.
  4. Calculate in degree mode and round once, at the end.
  5. Check the hypotenuse is the longest side.
Example 1 — The hypotenuse from sin
The side opposite a \(38^\circ\) angle is \(12\) cm. Find the hypotenuse \(x\), correct to one decimal place.
Solution
Opposite 12 cm, hypotenuse xAngle 38 degrees at the bottom left, opposite side 12 cm, hypotenuse x in red. 38° 12 cm x

Opposite and hypotenuse, so use sin. This time \(x\) is underneath:

\(\sin 38^\circ\)\(=\)\(\dfrac{12}{x}\)
\(x \sin 38^\circ\)\(=\)\(12\)
\(x\)\(=\)\(\dfrac{12}{\sin 38^\circ}\)
\(\)\(=\)\(19.491\ldots\)
\(\)\(\approx\)\(\mathbf{19.5} \text{ cm}\)
x=12sin38°≈19.5

\(19.5 > 12\), as the hypotenuse must be.

Example 2 — The hypotenuse from cos, with minutes
The side adjacent to an angle of \(54^\circ 18'\) is \(9\) m. Find the hypotenuse \(y\), correct to one decimal place.
Solution
Adjacent 9 m, hypotenuse yAngle 54 degrees 18 minutes at the bottom left, adjacent side 9 m, hypotenuse y in red. 54°18′ 9 m y

Adjacent and hypotenuse, so use cos. \(18' = 0.3^\circ\).

\(\cos 54.3^\circ\)\(=\)\(\dfrac{9}{y}\)
\(y\)\(=\)\(\dfrac{9}{\cos 54.3^\circ}\)
\(\)\(=\)\(15.423\ldots\)
\(\)\(\approx\)\(\mathbf{15.4} \text{ m}\)
y=9cos54.3°≈15.4
Example 3 — An adjacent side from tan
The side opposite a \(29^\circ\) angle is \(6.3\) cm. Find the adjacent side \(n\), correct to one decimal place.
Solution
Opposite 6.3 cm, adjacent nAngle 29 degrees at the bottom left, opposite side 6.3 cm, adjacent side n in red. 29° n 6.3 cm

Opposite and adjacent, so use tan. The unknown \(n\) is underneath:

\(\tan 29^\circ\)\(=\)\(\dfrac{6.3}{n}\)
\(n\)\(=\)\(\dfrac{6.3}{\tan 29^\circ}\)
\(\)\(=\)\(11.365\ldots\)
\(\)\(\approx\)\(\mathbf{11.4} \text{ cm}\)
n=6.3tan29°≈11.4

A small angle has a long adjacent side, so \(n > 6.3\) makes sense.

Example 4 — A cable, then a second side
A cable runs from the top of a \(21\) m mast to the ground, making \(47^\circ\) with the ground. Find
(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.
Solution
Cable from the top of a mastA 21 m mast. A cable c, in red, runs from its top to the ground at 47 degrees. The distance g along the ground is also marked. 47° 21 m c g

(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:

\(\tan 47^\circ\)\(=\)\(\dfrac{21}{g}\)
\(g\)\(=\)\(\dfrac{21}{\tan 47^\circ}\)
\(\)\(=\)\(19.582\ldots\)
\(\)\(\approx\)\(\mathbf{19.6} \text{ m}\)
g=21tan47°≈19.6

Common pitfalls

Do not multiply out of habit. \(\sin 35^\circ = \dfrac{8}{x}\) gives \(x = \dfrac{8}{\sin 35^\circ} \approx 13.9\), not \(8 \sin 35^\circ \approx 4.6\).
Divide by the whole ratio. It is \(\dfrac{8}{\sin 35^\circ}\): work out \(\sin 35^\circ\) first, or use brackets on the calculator.
Check against the hypotenuse. If the answer should be the longest side but comes out shorter than a known side, the rearrangement went the wrong way.
Use unrounded values in a second step. Rounding the first answer and then using it again can change the last decimal place.

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.

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