Trigonometry in right-angled triangles – Year 9 maths lesson to download as PDF

Worked example 1: using sine

A right-angled triangle has an angle of 35° and a hypotenuse of 10 cm. Find the side opposite the angle.

We know \(H\) and want \(O\), so use sin: \(\sin 35^\circ = \dfrac{x}{10}\), \(x = 10 \sin 35^\circ = 5.74\) cm to 3 s.f.

Worked example 2: using tangent

Angle 50°, adjacent 6 cm. Find the opposite side.

We know \(A\) and want \(O\): \(\tan 50^\circ = \dfrac{x}{6}\), \(x = 6 \tan 50^\circ = 7.15\) cm.

Worked example 3: unknown on the bottom

Angle 25°, opposite 7 cm. Find the hypotenuse.

\(\sin 25^\circ = \dfrac{7}{x}\). The unknown is on the bottom, so swap: \(x = \dfrac{7}{\sin 25^\circ} = 16.6\) cm.

Worked example 4: inverse sine

Opposite 3 cm, hypotenuse 5 cm. Find the angle.

\(\sin\theta = \dfrac{3}{5} = 0.6\), so \(\theta = \sin^{-1}(0.6) = 36.9^\circ\).

Worked example 5: inverse cosine

Adjacent 4 cm, hypotenuse 7 cm. Find the angle.

\(\cos\theta = \dfrac{4}{7}\), \(\theta = \cos^{-1}\left(\dfrac{4}{7}\right) = 55.2^\circ\).

Worked example 6: inverse tangent

Opposite 5 cm, adjacent 8 cm. Find the angle.

\(\tan\theta = \dfrac{5}{8} = 0.625\), \(\theta = \tan^{-1}(0.625) = 32.0^\circ\).

Worked example 7: elevation, ladder, isosceles triangle

(a) A tower is 30 m tall and you stand 45 m from its base: find the angle of elevation. (b) A 6 m ladder makes an angle of 70° with the ground: find the height reached and the distance of the foot from the wall. (c) An isosceles triangle has base 10 cm and equal sides 13 cm: find a base angle.

(a) The angle of elevation is measured upwards from the horizontal. \(\tan\theta = \dfrac{30}{45}\), so \(\theta = \tan^{-1}(0.667) = 33.7^\circ\). (b) Height \(6 \sin 70^\circ = 5.64\) m and foot \(6 \cos 70^\circ = 2.05\) m from the wall; check with Pythagoras: \(5.64^2 + 2.05^2 \approx 36\). (c) Split it down the middle to get a right-angled triangle with adjacent 5 and hypotenuse 13: the base angle is \(\cos^{-1}\left(\dfrac{5}{13}\right) = 67.4^\circ\).

Worked example 8: finding the hypotenuse

A ramp rises 1.2 m at an angle of \(15^\circ\) to the ground. Find the length of the ramp.

The rise is opposite the angle and we need the hypotenuse, so use sine: \(\sin15^\circ=\dfrac{1.2}{L}\). Rearranging, \(L=1.2\div\sin15^\circ=4.64\) m (to 2 decimal places).

Worked example 9: the angle of depression

From the top of a cliff 50 m high, the angle of depression of a boat is \(18^\circ\). How far is the boat from the base of the cliff?

The angle of depression equals the angle of elevation from the boat to the top, by alternate angles. \(\tan18^\circ=\dfrac{50}{d}\), so \(d=50\div\tan18^\circ=153.9\) m.

Worked example 10: two triangles

In triangle \(ABC\), angle \(B=90^\circ\), \(AB=8\) cm and angle \(A=35^\circ\). Find \(BC\) and then \(AC\).

\(BC=8\tan35^\circ=5.60\) cm. \(AC=8\div\cos35^\circ=9.77\) cm. Check with Pythagoras: \(\sqrt{8^2+5.6^2}=\sqrt{95.36}=9.77\).

Worked example 11: choosing the right ratio

A kite string is 25 m long and the kite is 18 m above the ground. Find the angle the string makes with the ground.

We know the opposite (18 m) and the hypotenuse (25 m), so use sine. \(\sin\theta=\dfrac{18}{25}=0.72\), so \(\theta=\sin^{-1}0.72=46.1^\circ\).

Worked example 12: a complete solution

A ramp rises 1.2 m over a horizontal distance of 9 m. Find the angle of the ramp to the horizontal, to 1 decimal place.

First sketch the right-angled triangle and mark the angle \(\theta\) at the bottom of the ramp. The side of 1.2 m is opposite \(\theta\) and the side of 9 m is adjacent to it. The hypotenuse is not known and not needed, so the ratio to use is tangent. Write \(\tan\theta=\dfrac{1.2}{9}=0.1333\ldots\) Then use the inverse function: \(\theta=\tan^{-1}(0.1333\ldots)=7.594\ldots^\circ\), which is \(7.6^\circ\) to 1 decimal place. A gentle slope should have a small angle, so the answer is sensible.