Pythagoras and trigonometry in right-angled triangles – Year 11 GCSE maths revision notes to download as PDF

Worked example 1: the hypotenuse

Legs 4 cm and 7 cm. Find the hypotenuse.

\(c^2=16+49=65\), so \(c=\sqrt{65}=8.06\) cm (2 d.p.).

Worked example 2: a ladder

A 5 m ladder has its foot 1.4 m from a wall. How high does it reach?

Height \(=\sqrt{5^2-1.4^2}=\sqrt{25-1.96}=\sqrt{23.04}=4.8\) m.

Worked example 3: finding a side

Angle \(40^\circ\), hypotenuse 8 cm. Find the adjacent side.

\(\cos40^\circ=\dfrac{x}{8}\), so \(x=8\cos40^\circ=6.13\) cm.

Worked example 4: finding an angle

Opposite 7, adjacent 9. Find the angle.

Use the inverse function. \(\tan\theta=\dfrac79\), so \(\theta=\tan^{-1}\dfrac79=37.9^\circ\). Check that your calculator is in degree mode.

Worked example 5: exact values and an isosceles triangle

(a) A right-angled triangle has hypotenuse 12 and an angle of \(60^\circ\): find the other sides exactly. (b) Find the height, area and base angle of an isosceles triangle with sides \(10,10,12\).

(a) Adjacent \(=12\cos60^\circ=6\) and opposite \(=12\sin60^\circ=6\sqrt3\). (b) Split it down the middle: half the base is 6, height \(=\sqrt{100-36}=8\), area \(=\tfrac12\times12\times8=48\) and each base angle \(=\cos^{-1}\dfrac6{10}=53.1^\circ\).

Worked example 6 (Higher): cuboid and pyramid

Find the space diagonal of a cuboid \(3\times4\times12\) and the angle it makes with the base. For a square-based pyramid with base 8 and height 6, find the slant height and a sloping edge.

The diagonal across the base is \(\sqrt{3^2+4^2}=5\). The space diagonal uses a second right-angled triangle with sides 5 and 12: \(\sqrt{5^2+12^2}=13\). The angle with the base is \(\tan^{-1}\dfrac{12}{5}=67.4^\circ\). For the pyramid, the slant height of a face is \(\sqrt{6^2+4^2}=\sqrt{52}=7.21\) and a sloping edge is \(\sqrt{6^2+(4\sqrt2)^2}=\sqrt{68}=8.25\).

Worked example 7: distance and bearing

A ship leaves port and sails 16 km due east, then 12 km due north. How far is it from the port and on what bearing?

The two legs form a right angle, so the distance is \(\sqrt{16^2+12^2}=\sqrt{400}=20\) km. The angle at the port between east and the direct line is \(\tan^{-1}\dfrac{12}{16}=36.9^\circ\). A bearing is measured clockwise from north, so the bearing is \(90^\circ-36.9^\circ=053.1^\circ\).

Worked example 8: Pythagoras in an isosceles triangle

An isosceles triangle has base 10 cm and equal sides 13 cm. Find its height and area.

The height meets the base at its midpoint, forming a right-angled triangle with hypotenuse 13 and base 5. Height \(=\sqrt{13^2-5^2}=12\) cm. Area \(=\tfrac12\times10\times12=60\) cm\(^2\). Dropping a perpendicular height is the standard way to use Pythagoras in isosceles and equilateral triangles.

Worked example 9: Choosing the correct trigonometric ratio

In a right-angled triangle the side opposite angle \(x\) is 6 cm and the hypotenuse is 10 cm. Find \(x\).

Label the sides: opposite is 6 and hypotenuse is 10. These two appear in the sine ratio (SOH). \(\sin x=\dfrac6{10}=0.6\), so \(x=\sin^{-1}0.6=36.9^\circ\). If you know the opposite and the adjacent use TOA, and if you know the adjacent and the hypotenuse use CAH.

Worked example 10: Angle of depression

From the top of a cliff 40 m high, the angle of depression of a boat is \(22^\circ\). How far is the boat from the foot of the cliff?

The angle of elevation of the top from the boat equals the angle of depression, \(22^\circ\). Distance \(=\dfrac{40}{\tan22^\circ}=\dfrac{40}{0.4040}=99.0\) m. Always draw the horizontal line first and measure the depression angle downwards from it.

Worked example 11: Distance between two points

Find the distance between \(A(1,2)\) and \(B(7,10)\).

Make a right-angled triangle with horizontal side \(7-1=6\) and vertical side \(10-2=8\). \(AB=\sqrt{6^2+8^2}=\sqrt{100}=10\). This is the distance formula, and it is just Pythagoras\' theorem applied to the coordinate grid.

Worked example 12: Two triangles in one diagram

In the diagram \(AB=12\) cm, angle \(BAC=90^\circ\), angle \(ABC=35^\circ\) and \(D\) is on \(BC\) with \(AD\perp BC\). Find \(AD\).

First find \(AC\) in triangle \(ABC\): \(AC=12\tan35^\circ=8.40\) cm. Then use triangle \(ABD\) (right-angled at \(D\)): \(AD=AB\sin35^\circ=12\times0.5736=6.88\) cm. Choose the triangle that contains the side you want and two pieces of information.

Worked example 13: A 3D problem (Higher)

A vertical pole 6 m tall stands at the corner \(A\) of a rectangular field 8 m by 15 m. Find the angle between the line from the top of the pole to the opposite corner and the ground.

The diagonal of the field is \(\sqrt{8^2+15^2}=17\) m. The angle is \(\tan^{-1}\dfrac6{17}=19.4^\circ\). In 3D problems always find the length in the horizontal plane first, then use a right-angled triangle that is vertical.

Worked example 14: Pythagoras to check a diagram

A rectangular gate is 1.2 m wide and 0.9 m tall. How long is the diagonal brace?

The brace is the hypotenuse: \(\sqrt{1.2^2+0.9^2}=\sqrt{1.44+0.81}=\sqrt{2.25}=1.5\) m. The sides \(0.9\), \(1.2\) and \(1.5\) are a 3, 4, 5 triangle scaled by 0.3.