Solving right-angled triangles
The floodlight pillar beside your school pitch is far too tall to climb with a tape measure. But you can measure a distance along the ground and an angle up to the top. Is that enough to find the height? By the end of this page you will see that it is.
Naming the sides and the three ratios
In a right-angled triangle, the sides are named from the point of view of one of the acute angles. Call that angle $\theta$.
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The hypotenuse is the longest side. It is always opposite the right angle.
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The opposite side is directly across from $\theta$.
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The adjacent side is the one next to $\theta$ (that is not the hypotenuse).
If you change to the other acute angle, "opposite" and "adjacent" swap. The hypotenuse never changes.
The three trigonometric ratios compare two sides of the triangle:
A handy memory aid is SOH CAH TOA: Sin = Opposite over Hypotenuse, Cos = Adjacent over Hypotenuse, Tan = Opposite over Adjacent.
Why do these ratios work? Any two right-angled triangles with the same acute angle are similar, so their sides are in the same proportion. The ratio opposite ÷ hypotenuse therefore depends only on the angle, not on the size of the triangle. That is why a calculator can store $\sin\theta$ for every angle.
In a right-angled triangle $ABC$, the right angle is at $B$, $AB = 6 \text{ cm}$ and $BC = 8 \text{ cm}$. Calculate the length of $AC$, and then write down $\sin A$,…
In a right-angled triangle $ABC$, the right angle is at $B$, $AB = 6 \text{ cm}$ and $BC = 8 \text{ cm}$. Calculate the length of $AC$, and then write down $\sin A$, $\cos A$ and $\tan A$ as fractions.
Using the calculator
Before you press a single key, make sure the calculator is in degree mode (a small D or DEG shows on the screen). In radian mode every answer will be wrong.
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To find a ratio from an angle, press the function key and then the angle. For example $\sin 50^\circ = 0.7660$ (to four decimal places).
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To find an angle from a ratio, use the inverse function. Press SHIFT and then sin, cos or tan to get $\sin^{-1}$, $\cos^{-1}$ or $\tan^{-1}$. For example, if $\cos A = 0.25$ then $A = \cos^{-1}(0.25) = 75.5^\circ$ (to one decimal place).
Read $\sin^{-1}(0.5)$ as "the angle whose sine is 0.5". It is not $\frac{1}{\sin 0.5}$.
Use a calculator to find the following. Give answers correct to two decimal places for parts a) and b), and correct to one decimal place for parts c) and d). a)…
Use a calculator to find the following. Give answers correct to two decimal places for parts a) and b), and correct to one decimal place for parts c) and d).
- a) $\sin 35^\circ$
- b) $\tan 62^\circ$
- c) the angle $A$ such that $\cos A = 0.4$
- d) the angle $B$ such that $\tan B = 1.5$
Finding a missing side
Choose the ratio that contains the side you know and the side you want. Ignore the third side.
When the unknown ends up on the bottom of the fraction, rearrange before calculating.
In a right-angled triangle, the hypotenuse is $12 \text{ cm}$ and one of the acute angles is $40^\circ$. a) Calculate the length of the side opposite the $40^\circ$…
In a right-angled triangle, the hypotenuse is $12 \text{ cm}$ and one of the acute angles is $40^\circ$.
- a) Calculate the length of the side opposite the $40^\circ$ angle, correct to one decimal place.
- b) Calculate the length of the side adjacent to the $40^\circ$ angle, correct to one decimal place.
Finding a missing angle
If you know two sides, pick the ratio that uses them, work out the fraction, then use the inverse function.
Once you know one acute angle, the other is $90^\circ$ minus it, since the angles of a triangle add up to $180^\circ$ [Theorem 4].
A ladder $5.5 \text{ m}$ long leans against a vertical wall of a house in Galway. The foot of the ladder is $1.8 \text{ m}$ from the wall on level ground. a) Calculate…
A ladder $5.5 \text{ m}$ long leans against a vertical wall of a house in Galway. The foot of the ladder is $1.8 \text{ m}$ from the wall on level ground.
- a) Calculate the angle the ladder makes with the ground, correct to one decimal place.
- b) Calculate how high up the wall the ladder reaches, correct to two decimal places.
Angles of elevation and depression
Real problems describe an angle measured from the horizontal.
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The angle of elevation is the angle above the horizontal when you look up at something.
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The angle of depression is the angle below the horizontal when you look down at something.
The two angles are equal, because they are alternate angles between parallel horizontal lines [Theorem 5, alternate angles]. Always draw a sketch and mark the right-angled triangle.
From a point on level ground $45 \text{ m}$ from the base of the Spire in Dublin, the angle of elevation of the top of the Spire is $67.6^\circ$. a) Calculate the height…
From a point on level ground $45 \text{ m}$ from the base of the Spire in Dublin, the angle of elevation of the top of the Spire is $67.6^\circ$.
- a) Calculate the height of the Spire, correct to the nearest metre.
- b) Calculate the straight-line distance from the point on the ground to the top of the Spire, correct to the nearest metre.
A lifeguard stands on a cliff $32 \text{ m}$ above sea level near the Cliffs of Moher. She sees a swimmer at an angle of depression of $18^\circ$. a) Draw a sketch in…
A lifeguard stands on a cliff $32 \text{ m}$ above sea level near the Cliffs of Moher. She sees a swimmer at an angle of depression of $18^\circ$.
- a) Draw a sketch in your copy and mark the angle of depression and the angle of elevation from the swimmer.
- b) Calculate the horizontal distance from the swimmer to the base of the cliff, correct to the nearest metre.
Multi-step problems
Harder problems contain two right-angled triangles, or a rectangle with a diagonal. Split the diagram into separate right-angled triangles and solve one at a time.
Two points $A$ and $B$ lie on level ground on opposite sides of a tower $PQ$, where $Q$ is the base of the tower and $A$, $Q$, $B$ lie in a straight line. The angle of…
Two points $A$ and $B$ lie on level ground on opposite sides of a tower $PQ$, where $Q$ is the base of the tower and $A$, $Q$, $B$ lie in a straight line. The angle of elevation of the top $P$ from $A$ is $28^\circ$ and from $B$ is $41^\circ$. The distance $AB$ is $120 \text{ m}$.
- a) Let $h$ be the height of the tower. Write $AQ$ and $BQ$ in terms of $h$.
- b) Hence calculate $h$, correct to one decimal place.
A rectangle $PQRS$ has $PQ = 14 \text{ cm}$ and $QR = 9 \text{ cm}$. The diagonal $PR$ is drawn. a) Calculate the angle $RPQ$, correct to one decimal place. b) Calculate…
A rectangle $PQRS$ has $PQ = 14 \text{ cm}$ and $QR = 9 \text{ cm}$. The diagonal $PR$ is drawn.
- a) Calculate the angle $RPQ$, correct to one decimal place.
- b) Calculate the length of the diagonal $PR$ in two different ways, and verify that the answers agree.
- c) A point $T$ lies on $PQ$ such that angle $TRQ = 30^\circ$. Calculate the length $TQ$, correct to two decimal places.
A pole stands on top of a building. From a point $50 \text{ m}$ from the base of the building, on level ground, the angle of elevation of the bottom of the pole is…
A pole stands on top of a building. From a point $50 \text{ m}$ from the base of the building, on level ground, the angle of elevation of the bottom of the pole is $35^\circ$ and the angle of elevation of the top of the pole is $39^\circ$.
- a) Calculate the height of the pole, correct to one decimal place.
- b) Explain why it would be incorrect to calculate the height of the pole as $50 \tan 4^\circ$.
Summary
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Label the sides from the chosen angle: hypotenuse, opposite, adjacent.
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$\sin\theta = \frac{\text{opp}}{\text{hyp}}$, $\cos\theta = \frac{\text{adj}}{\text{hyp}}$, $\tan\theta = \frac{\text{opp}}{\text{adj}}$ (SOH CAH TOA).
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Use degree mode. To find a side, choose the ratio containing the known and unknown sides. To find an angle, use $\sin^{-1}$, $\cos^{-1}$ or $\tan^{-1}$.
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Angles of elevation and depression are measured from the horizontal and are equal to each other (alternate angles). Always draw a sketch.
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In multi-step problems, split the diagram into right-angled triangles, and do not subtract angles.
Back to the floodlight pillar: standing $20$ m away, you measure an angle of elevation of $52^\circ$. Then $h = 20 \tan 52^\circ = 20 \times 1.2799 \approx 25.6$ m above your eye level. Add your eye height to get the full height of the pillar.