Split the 140° angle at A into a right angle and a 50° angle.
You can then form a right angled triangle using the 50° angle and the side AB as the hypotenuse.
Label the side adjacent to the 50° as a and the side opposite as b.

cie-igcse-2020-may-jun-p4-tz2-q4b-1-ma

You can find the values of a and b using the trig ratios: $\sin (50) = \frac{Opposite}{Hypotenuse}$ and $\cos (50) = \frac{Adjacent}{Hypotenuse} .$

$\begin{array}{rcl} \sin (50) & = & \frac{b}{400} \\ \cos (50) & = & \frac{a}{400} \end{array}$

1 mark for each correct equation [2]

Rearrange both equations by multiplying both sides by 400.
Leave the values for a and b in exact form to avoid rounding errors.

$\begin{array}{rcl} a & = & 400 \cos (50) \\ b & = & 400 \sin (50) \end{array}$

You can then form a right angled triangle using AD as the hypotenuse and the right angle given in the diagram.
The horizontal length can be found by adding a to 350 m.
The vertical length can be found by subtracting b from 450 m.

cie-igcse-2020-may-jun-p4-tz2-q4b-2-ma

Correct expression for horizontal length [1]
Correct expression for vertical length [1]

You can then use Pythagoras' theorem $c^{2} = a^{2} + b^{2}$ to find the length of AD.

$A D^{2} = {(350 + 400 \cos (50))}^{2} + {(450 - 400 \sin (50))}^{2}$

[1]

Square root both sides of the expression.

$A D = \sqrt{{(350 + 400 \cos (50))}^{2} + {(450 - 400 \sin (50))}^{2}}$

Type the expression in your calculator.

$A D = 623.8625 . . .$

Round to 3 significant figures.

$A D = 624 m$ [1]

Right-Angled Triangles - Pythagoras & Trigonometry (CIE IGCSE Maths: Extended)

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