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Pythagoras & Trigonometry

10 exam-style questions on Pythagoras’ theorem, right-angled trigonometry and exact values.

3 Easy 4 Medium 3 Hard
Q1Easy3 marks

A right-angled triangle has legs \( 5 \) cm and \( 12 \) cm. Find the hypotenuse.

\( \sqrt{25 + 144} = \sqrt{169} = 13 \) cm.

Q2Easy3 marks

In a right-angled triangle, opposite \( = 4 \), adjacent \( = 3 \). Find \( \sin \theta \) and \( \cos \theta \).

Hypotenuse \( = 5 \). \( \sin \theta = \dfrac{4}{5} \), \( \cos \theta = \dfrac{3}{5} \).

Q3Easy3 marks

Write down the exact value of \( \sin 30^\circ \) and \( \tan 45^\circ \).

\( \sin 30^\circ = \dfrac{1}{2} \), \( \tan 45^\circ = 1 \).

Q4Medium4 marks

A ladder \( 5 \) m long leans against a wall. The foot is \( 1.5 \) m from the wall. How far up the wall does it reach?

\( h = \sqrt{25 - 2.25} = \sqrt{22.75} \approx 4.77 \) m.

Q5Medium4 marks

In a right-angled triangle, the angle is \( 35^\circ \) and the adjacent side is \( 8 \) cm. Find the opposite side to 1 d.p.

\( \tan 35^\circ = \dfrac{\text{opp}}{8} \) ⇒ opp \( = 8 \tan 35^\circ \approx 5.6 \) cm.

Q6Medium4 marks

Find the size of angle \( \theta \) if \( \sin \theta = 0.6 \) and \( \theta \) is acute. Give your answer to the nearest degree.

\( \theta = \sin^{-1}(0.6) \approx 37^\circ \).

Q7Medium4 marks

A triangle has sides \( 7 \) cm, \( 24 \) cm and \( 25 \) cm. Is it right-angled? Justify.

\( 7^2 + 24^2 = 49 + 576 = 625 = 25^2 \). Yes, by converse of Pythagoras.

Q8Hard5 marks

An isosceles right-angled triangle has equal sides of length \( 6 \) cm. Find the exact length of the hypotenuse.

\( \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} \) cm.

Q9Hard5 marks

From a point \( 40 \) m from the base of a building, the angle of elevation to the top is \( 28^\circ \). Find the height of the building to 1 d.p.

\( h = 40 \tan 28^\circ \approx 21.3 \) m.

Q10Hard5 marks

Write down the exact value of \( \cos 60^\circ \) and use it to find the exact length of the adjacent side when the hypotenuse is \( 10 \) cm and the angle is \( 60^\circ \).

\( \cos 60^\circ = \dfrac{1}{2} \). Adjacent \( = 10 \times \dfrac{1}{2} = 5 \) cm.