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Year 9 Maths Core Trigonometry

Finding Unknown Side

20 practice questions 2 video lessons Theory + worked examples

Once an angle and one side of a right-angled triangle are known, a trigonometric ratio finds another side. In this NSW Year 9 Core maths subtopic the unknown is always on the top of the ratio, so the last step is a multiplication, as in \(x = 12\cos 35^\circ\). It comes under NESA outcome MA5‑TRG‑C‑01.

You will label the sides from the given angle, choose sine, cosine or tangent from the two sides involved, substitute and multiply, and round to one or two decimal places. You will also work with angles given in degrees and minutes, triangles described only by their vertices, and worded problems such as the height of a boat's mast, the rise of a wheelchair ramp and how far up a wall a ladder reaches. It is part of Year 9 Core Trigonometry.

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Practice questions

Every question with a fully worked solution.

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Watch 2 video(s)
  • Finding Unknown Side - Video - Finding Missing Sides Watch
  • Finding Unknown Side - Video - Writing Trig Equations to Find a Missing Side of a Right Triangle Watch
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Theory

To find an unknown side of a right-angled triangle with trigonometry, label the sides from the given angle, choose the ratio that uses the known side and the wanted side, and multiply. This Year 9 maths page covers sin, cos and tan, angles in degrees and minutes, triangles named by their vertices, and worded problems.

When one angle (besides the right angle) and one side of a right-angled triangle are known, the other sides can be found with a trigonometric ratio.

The sides are labelled from the given angle: the hypotenuse faces the right angle, the opposite faces the angle, and the adjacent is next to it.

Choose the ratio that contains BOTH the side you know and the side you want. On this page the unknown side is on top of the fraction, so the last step is a multiplication.

Angles can be given in degrees and minutes. One degree is \(60\) minutes, so \(48^\circ 24' = 48.4^\circ\).

Known side and wanted sideA right-angled triangle with angle theta. The adjacent side is known and the opposite side x, in red, is wanted. Those two sides point to the tan ratio. θ known: adjacent want: x hypotenuse
Look at the two sides in the question: the one you know and the one you want. Adjacent and opposite point to \(\tan\).
Which two sides go with each ratioThree small triangles. For sin the opposite side and the hypotenuse are red; for cos the adjacent side and the hypotenuse; for tan the opposite and the adjacent sides. sin opp, hyp cos adj, hyp tan opp, adj
Each ratio uses exactly two sides (red). Pick the one that contains both the known side and the wanted side.

The three ratios:

\[\sin \theta = \dfrac{\text{opp}}{\text{hyp}} \qquad \cos \theta = \dfrac{\text{adj}}{\text{hyp}} \qquad \tan \theta = \dfrac{\text{opp}}{\text{adj}}\]
tanθ=oppadj

With the unknown on top, multiply both sides by the bottom:

\[\text{opp} = \text{hyp} \times \sin \theta \qquad \text{adj} = \text{hyp} \times \cos \theta \qquad \text{opp} = \text{adj} \times \tan \theta\]
opp=hyp×sinθ
Round once, at the end. Keep the full calculator value until the last line. If one answer is used again, as when finding two sides for a perimeter or area, use the unrounded value.
Isosceles triangles. The line from the apex to the middle of the base cuts it into two right-angled triangles. It halves the apex angle and the base.

How to find an unknown side

  1. Sketch the triangle and mark the right angle, the given angle, the known side and the wanted side.
  2. Label hyp, opp and adj from the given angle.
  3. Choose the ratio that uses the known side and the wanted side.
  4. Substitute and multiply to get the unknown on its own. Change any minutes into degrees first.
  5. Round as asked, and check a shorter side is shorter than the hypotenuse.
Example 1 — Using sin
A right-angled triangle has a hypotenuse of \(18\) cm and an angle of \(35^\circ\). Find the side opposite that angle, correct to one decimal place.
Solution
Hypotenuse 18 cm, angle 35 degreesAngle 35 degrees at the bottom left, hypotenuse 18 cm, and the side x opposite the angle in red. 35° x 18 cm

From the \(35^\circ\) angle the sides are the opposite \(x\) and the hypotenuse \(18\), so use sin. The unknown is on top, so multiply:

\(\sin 35^\circ\)\(=\)\(\dfrac{x}{18}\)
\(x\)\(=\)\(18 \sin 35^\circ\)
\(\)\(=\)\(10.324\ldots\)
\(\)\(\approx\)\(\mathbf{10.3} \text{ cm}\)
x=18sin35°≈10.3

\(10.3 < 18\), as a shorter side must be.

Example 2 — A triangle named by its vertices
In \(\triangle KLM\), \(\angle KLM = 90^\circ\), \(\angle LKM = 62^\circ\) and \(KM = 26\) m. Find \(KL\), correct to one decimal place.
Solution

The right angle is at \(L\), so \(KM\) is the hypotenuse. From the angle at \(K\), \(KL\) runs to the right angle: it is the adjacent side. Use cos:

Triangle KLMTriangle KLM with the right angle at L, angle 62 degrees at K, KM 26 m as the hypotenuse, and KL, in red, along the bottom adjacent to K. 62° 26 m K L M
\(\cos 62^\circ\)\(=\)\(\dfrac{KL}{26}\)
\(KL\)\(=\)\(26 \cos 62^\circ\)
\(\)\(=\)\(12.206\ldots\)
\(\)\(\approx\)\(\mathbf{12.2} \text{ m}\)
KL=26cos62°≈12.2

Sketch the triangle from the words first; it makes the labelling obvious.

Example 3 — Using tan, with minutes
The side adjacent to an angle of \(48^\circ 24'\) is \(11\) cm. Find the opposite side \(p\), correct to one decimal place.
Solution
Adjacent 11 cm, angle 48 degrees 24 minutesAngle 48 degrees 24 minutes at the bottom left, adjacent side 11 cm, and the opposite side p in red. 48°24′ 11 cm p

Opposite and adjacent: no hypotenuse, so use tan. Change the minutes first: \(24' = \dfrac{24}{60} = 0.4^\circ\).

\(\tan 48.4^\circ\)\(=\)\(\dfrac{p}{11}\)
\(p\)\(=\)\(11 \tan 48.4^\circ\)
\(\)\(=\)\(12.389\ldots\)
\(\)\(\approx\)\(\mathbf{12.4} \text{ cm}\)
p=11tan48.4°≈12.4

Or enter \(48^\circ 24'\) directly with the degrees-minutes key.

Example 4 — A worded problem
A \(6\) m ladder leans against a vertical wall. It makes an angle of \(18^\circ\) with the WALL. How far is the foot of the ladder from the wall, correct to two decimal places?
Solution
Ladder leaning against a wallA 6 m ladder leans against a vertical wall and makes an angle of 18 degrees with the wall at the top. The distance d from the foot of the ladder to the wall is marked in red along the ground. 18° 6 m d wall

The angle is at the top, between the ladder and the wall. From there, the ladder is the hypotenuse and the distance \(d\) along the ground is opposite. Use sin:

\(\sin 18^\circ\)\(=\)\(\dfrac{d}{6}\)
\(d\)\(=\)\(6 \sin 18^\circ\)
\(\)\(=\)\(1.854\ldots\)
\(\)\(\approx\)\(\mathbf{1.85} \text{ m}\)
d=6sin18°≈1.85

If the angle had been measured from the ground, \(d\) would be the adjacent side instead.

Common pitfalls

Choose the ratio from the sides, not from habit. Write down which two sides are involved before choosing sin, cos or tan.
Minutes are sixtieths, not hundredths. \(15^\circ 30'\) is \(15.5^\circ\), not \(15.3^\circ\).
Check where the angle is measured from. An angle with the wall and an angle with the ground are different angles. Label the sides from the angle you are given.
A short side cannot beat the hypotenuse. If your opposite or adjacent side comes out longer than the hypotenuse, the wrong ratio was used.

Frequently asked questions

How do you find a missing side with trigonometry?

Label the sides from the given angle, choose the ratio that uses the known side and the wanted side, substitute, and solve. If the unknown is on top, multiply.

How do I know whether to use sin, cos or tan?

Look at the two sides in the question. Opposite and hypotenuse means sin, adjacent and hypotenuse means cos, opposite and adjacent means tan.

How do you use degrees and minutes in trigonometry?

Divide the minutes by \(60\) and add them to the degrees, or use the degrees-minutes key on the calculator. \(48^\circ 24' = 48.4^\circ\).

When do you multiply and when do you divide?

If the unknown side is on top of the ratio, multiply. If it is on the bottom, divide. That second case is covered in Finding a Side (Unknown Denominator).