You know an angle and the length of the side opposite for the right angled triangle ABC so you can use the sin ratio to find the length of the hypotenuse AC.

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$\begin{array}{rcl} \sin (60) & = & \frac{12}{y} \end{array}$

[1]

Multiply both sides by y and divide both sides by sin(60). Leave your answer as an exact value.

$\begin{array}{rcl} y & = & \frac{12}{\sin (60)} \\ = & \frac{12}{(\frac{\sqrt{3}}{2})} \\ = & 12 \times \frac{2}{\sqrt{3}} \\ = & \frac{24}{\sqrt{3}} \end{array}$

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The angle in the rectangle at C is 90°.

$\begin{array}{rcl} Angle A C D & = & 90 ° - 60 ° \\ = & 30 ° \end{array}$

You now know an angle and the length of the adjacent side for the right angled triangle ACE so you can use the cos ratio to find the length of the hypotenuse CE.

4-10-edexcel-gcse-vhq2-2

$\begin{array}{rcl} \cos (30) & = & \frac{(\frac{24}{\sqrt{3}})}{x} \end{array}$

[1]

Multiply both sides by x and divide both sides by cos(30).

$\begin{array}{rcl} x & = & \frac{(\frac{24}{\sqrt{3}})}{\cos (30)} \\ = & \frac{(\frac{24}{\sqrt{3}})}{(\frac{\sqrt{3}}{2})} \\ = & \frac{24}{\sqrt{3}} \times \frac{2}{\sqrt{3}} \\ = & \frac{48}{3} \\ = & 16 \end{array}$

[1]

Remember to include the units for the length.

$C E = 16 cm$ [1]

Right-Angled Triangles: Pythagoras & Trigonometry (Cambridge O Level Maths)

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