Finding a Side (Unknown Denominator)
Sometimes the unknown side ends up in the denominator, as in \(\sin 40^\circ = \frac{8}{x}\), so the equation is rearranged to \(x = \frac{8}{\sin 40^\circ}\). This NSW Year 9 Advanced maths subtopic practises this step (NESA outcome MA5‑TRG‑C‑01).
You will find hypotenuses and other sides where the known side sits on top of the ratio, check that each answer is sensible, and solve problems about a support cable, a building and a zip-line. You will find perimeters, work across two triangles that share a side, and use a formula for the perimeter in terms of \(\sin\theta\) and \(\tan\theta\). One problem follows a plane coming in to land. It is part of Year 9 Advanced Trigonometry.
Theory
When the unknown side is on the bottom of a trig ratio, it swaps places with the ratio: \(x = \dfrac{10}{\sin\theta}\). This Year 9 Advanced maths page covers dividing by a ratio, angles in degrees and minutes, two triangles that share a side, sides written with a pronumeral and perimeter formulas.
Name the sides from the given angle \(\theta\): the hypotenuse (opposite the right angle), the opposite side and the adjacent side. Then choose the ratio that links the side you know with the side you want: \(\sin\theta = \dfrac{O}{H}\), \(\cos\theta = \dfrac{A}{H}\), \(\tan\theta = \dfrac{O}{A}\).
The unknown lands on the bottom (the denominator) when it is the hypotenuse and you know the opposite or adjacent side, or when it is the adjacent side and you know the opposite side.
Then the unknown and the ratio swap places: from \(\cos\theta = \dfrac{7}{x}\), multiply both sides by \(x\) to get \(x\cos\theta = 7\), then divide by \(\cos\theta\) to get \(x = \dfrac{7}{\cos\theta}\).
For an acute angle, \(\sin\theta\) and \(\cos\theta\) are less than \(1\). Dividing by a number less than \(1\) makes the result bigger, which is why the hypotenuse comes out longer than the side you started with.
Angles may be given in degrees and minutes: \(1^\circ = 60'\), so \(52^\circ 45' = 52.75^\circ\).
| You know | You want | Ratio | Rearranged |
|---|---|---|---|
| opposite | hypotenuse | \(\sin\theta = \dfrac{O}{H}\) | \(H = \dfrac{O}{\sin\theta}\) |
| adjacent | hypotenuse | \(\cos\theta = \dfrac{A}{H}\) | \(H = \dfrac{A}{\cos\theta}\) |
| opposite | adjacent | \(\tan\theta = \dfrac{O}{A}\) | \(A = \dfrac{O}{\tan\theta}\) |
How to find a side on the bottom of the ratio
- Label the sides from the given angle. Convert any minutes to a decimal of a degree.
- Choose the ratio containing the known side and the unknown side.
- Write the equation, for example \(\tan 52^\circ = \dfrac{9}{x}\).
- Swap: \(x = \dfrac{9}{\tan 52^\circ}\). Calculate in degree mode.
- Check that the hypotenuse is the longest side, then round as asked, with units.
Harder set-ups
- Both other sides: find each one from the given side, so a rounding error in one does not pass to the other. A perimeter is the sum of the three sides.
- A height inside a triangle: a perpendicular splits the triangle into two right-angled triangles that share it. Find lengths in each, then add the parts of the base.
- A side in two triangles: find the shared side in the first triangle and carry its full value into the second.
- Angles with parallel lines: a slope seen from the top at an angle below the horizontal makes the same angle with the ground (alternate angles).
- Pronumerals in the sides: multiply both sides by the denominator, collect the \(x\) terms, factorise and divide (Example 3).
- General formulas: write each side in terms of one side and \(\theta\), then add or substitute (Example 4).
(i) \(DF\)
(ii) \(EF\).
From \(F\): \(DE\) is opposite, \(DF\) is the hypotenuse and \(EF\) is adjacent. Both unknowns end up on the bottom of a fraction.
(i) Opposite and hypotenuse: sine.
| \(\sin 33^\circ 48'\) | \(=\) | \(\dfrac{11.5}{DF}\) |
| \(DF\) | \(=\) | \(\dfrac{11.5}{\sin 33^\circ 48'}\) |
| \(\) | \(=\) | \(20.6724\ldots\) |
| \(\therefore\ DF\) | \(\approx\) | \(\mathbf{20.7}\text{ cm}\) |
(ii) Use the given side again, not the rounded answer to (i). Opposite and adjacent: tangent.
| \(\tan 33^\circ 48'\) | \(=\) | \(\dfrac{11.5}{EF}\) |
| \(EF\) | \(=\) | \(\dfrac{11.5}{\tan 33^\circ 48'}\) |
| \(\) | \(=\) | \(17.1784\ldots\) |
| \(\therefore\ EF\) | \(\approx\) | \(\mathbf{17.2}\text{ cm}\) |
\(33^\circ 48' = 33.8^\circ\). Check: the hypotenuse \(DF\) is the longest side.
(i) the length of each rope
(ii) the distance \(AB\) between the pegs.
In each triangle the pole is opposite the angle. Each rope is a hypotenuse (sine); each ground distance is adjacent (tangent).
(i)
| \(r_1\) | \(=\) | \(\dfrac{2.4}{\sin 58^\circ} = 2.8300\ldots\) |
| \(r_2\) | \(=\) | \(\dfrac{2.4}{\sin 46^\circ} = 3.3363\ldots\) |
The ropes are about \(\mathbf{2.83}\) m and \(\mathbf{3.34}\) m.
(ii)
| \(\text{left part}\) | \(=\) | \(\dfrac{2.4}{\tan 58^\circ} = 1.4996\ldots\) |
| \(\text{right part}\) | \(=\) | \(\dfrac{2.4}{\tan 46^\circ} = 2.3176\ldots\) |
| \(AB\) | \(=\) | \(3.8173\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{3.82}\text{ m}\) |
Add the two full calculator values, then round once.
Adjacent and hypotenuse: cosine. The unknown is on the top and the bottom, so multiply out and collect the \(x\) terms.
| \(\cos 40^\circ\) | \(=\) | \(\dfrac{x + 4}{2x + 1}\) |
| \((2x + 1)\cos 40^\circ\) | \(=\) | \(x + 4\) |
| \(2x\cos 40^\circ - x\) | \(=\) | \(4 - \cos 40^\circ\) |
| \(x\,(2\cos 40^\circ - 1)\) | \(=\) | \(4 - \cos 40^\circ\) |
| \(x\) | \(=\) | \(\dfrac{4 - \cos 40^\circ}{2\cos 40^\circ - 1}\) |
| \(\) | \(=\) | \(6.0778\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{6.08}\) |
Check: the sides are about \(10.08\) and \(13.16\), and \(10.08 \div 13.16\) \(\approx 0.766\) \(= \cos 40^\circ\).
(i) Show that its perimeter is \(P\) \(= b\left(1 + \tan\theta + \dfrac{1}{\cos\theta}\right)\).
(ii) Find \(P\) when \(b = 12\) cm and \(\theta = 28^\circ\), correct to one decimal place.
(i) Write the other two sides in terms of \(b\) and \(\theta\).
| \(\tan\theta\) | \(=\) | \(\dfrac{\text{opp}}{b} \ \Rightarrow\ \text{opp} = b\tan\theta\) |
| \(\cos\theta\) | \(=\) | \(\dfrac{b}{\text{hyp}} \ \Rightarrow\ \text{hyp} = \dfrac{b}{\cos\theta}\) |
| \(P\) | \(=\) | \(b + b\tan\theta + \dfrac{b}{\cos\theta}\) |
| \(\) | \(=\) | \(b\left(1 + \tan\theta + \dfrac{1}{\cos\theta}\right)\) |
(ii) Substitute.
| \(P\) | \(=\) | \(12\left(1 + \tan 28^\circ + \dfrac{1}{\cos 28^\circ}\right)\) |
| \(\) | \(=\) | \(31.9713\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{32.0}\text{ cm}\) |
The hypotenuse needs division by \(\cos\theta\) because \(b\) is on the top of the cosine ratio and the hypotenuse is on the bottom.
Common pitfalls
Frequently asked questions
How do you find the hypotenuse with trigonometry?
If you know the opposite side, \(H = \dfrac{O}{\sin\theta}\). If you know the adjacent side, \(H = \dfrac{A}{\cos\theta}\).
Why do you divide when the unknown is on the bottom?
Multiplying both sides by \(x\) gives \(x\sin\theta = O\). To get \(x\) alone, divide both sides by \(\sin\theta\): \(x = \dfrac{O}{\sin\theta}\).
Why is the answer bigger than the side I started with?
Sine and cosine of an acute angle are less than \(1\), and dividing by a number less than \(1\) gives a larger result. That matches the hypotenuse being the longest side.
How do you use degrees and minutes in a trig ratio?
Divide the minutes by \(60\) and add them to the degrees, or use the degrees-minutes-seconds key: \(52^\circ 45' = 52.75^\circ\).
What if the side lengths are given with x in them?
Write the ratio as a fraction of the two expressions, multiply both sides by the denominator, collect the \(x\) terms on one side, factorise and divide.