Year 10 Maths - 10A Trigonometry

Finding Unknown Sides (in Denominator)

Q: 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 equals the known side divided by the ratio.

Q: 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.

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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 10A maths page revises the Year 9 method: rearrange so that the last step is a division, tell which case you are in, and see why the answer comes out bigger than the known side when it is the hypotenuse. It also covers sides written with \(x\) and multi-step problems.

This revises the Year 9 work. 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.

Same triangle shape, same ratio. Only the position of the unknown decides whether to multiply or to divide.

\(\sin 35^\circ \approx 0.57\) 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 = \frac{k}{\sin θ}

The pattern:

\[\text{unknown} = \dfrac{\text{known side}}{\text{the ratio}}\]

unknown = \frac{known side}{ratio}

The ratio reads	Unknown is	Last step
\(\cos \theta = \dfrac{x}{20}\)	on top	\(x = 20 \cos \theta\)
\(\cos \theta = \dfrac{20}{x}\)	underneath	\(x = \dfrac{20}{\cos \theta}\)

Letters instead of numbers. If the side opposite \(\theta\) is \(a\), the same rearrangement gives hypotenuse \(= \dfrac{a}{\sin \theta}\) and adjacent \(= \dfrac{a}{\tan \theta}\). Add these to \(a\) for a perimeter. When the sides are expressions in \(x\), multiply by the denominator, collect the \(x\) terms and factor out \(x\).

Speed and time. Once a length is found, time \(=\) distance \(\div\) speed. Convert the units to match first, for example km/h to m/min.

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

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.
Carry on if the question asks more: a second triangle, a perimeter, or a time from a speed. Use the unrounded length.

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 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 = \frac{12}{\sin 38 °} \approx 19.5

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

Example 2 — Sides written with \(x\)

In a right-angled triangle, the side adjacent to a \(50^\circ\) angle is \((x + 3)\) cm and the hypotenuse is \(2x\) cm. Find \(x\), correct to two decimal places.

Solution

Adjacent and hypotenuse, so use cos. Multiply both sides by the denominator, then collect the \(x\) terms:

\(\cos 50^\circ\)	\(=\)	\(\dfrac{x + 3}{2x}\)
\(2x \cos 50^\circ\)	\(=\)	\(x + 3\)
\(2x \cos 50^\circ - x\)	\(=\)	\(3\)
\(x(2\cos 50^\circ - 1)\)	\(=\)	\(3\)
\(x\)	\(=\)	\(\dfrac{3}{2\cos 50^\circ - 1}\)
\(\)	\(=\)	\(10.505\ldots\)
\(\)	\(\approx\)	\(\mathbf{10.51}\)

x = \frac{3}{2 \cos 50 ° - 1} \approx 10.51

Check: the hypotenuse is \(2x \approx 21.0\), longer than \(x + 3 \approx 13.5\).

Example 3 — An adjacent side from tan, with minutes

The side opposite an angle of \(29^\circ 24'\) is \(6.3\) cm. Find the adjacent side \(n\), correct to one decimal place.

Solution

Opposite and adjacent, so use tan. \(24' = \dfrac{24}{60}^\circ = 0.4^\circ\). The unknown \(n\) is underneath:

\(\tan 29.4^\circ\)	\(=\)	\(\dfrac{6.3}{n}\)
\(n\)	\(=\)	\(\dfrac{6.3}{\tan 29.4^\circ}\)
\(\)	\(=\)	\(11.180\ldots\)
\(\)	\(\approx\)	\(\mathbf{11.2} \text{ cm}\)

n = \frac{6.3}{\tan 29.4 °} \approx 11.2

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

(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 = \frac{21}{\tan 47 °} \approx 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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Finding Unknown Angles

Finding Unknown Sides (in Denominator)

📖 Theory

How to find a side when the unknown is underneath

Common pitfalls

Frequently asked questions

Theory