The diagram shows part of a map of a small town. Joel's house is located at point W, the childcare centre at point X, the park at point Y and the shopping mall at point Z. WZ = 4.6 km, YZ = 2.5 km and XY = 3.4 km The bearing of W from Y is 043.4° and the bearing of X from Y is 064.1°.

  1. ((a))

    Find the bearing of park Y from childcare centre X.

  2. ((b))

    Find the distance of park Y from Joel's house W.

  3. ((c))

    Find the bearing of the shopping mall Z from Joel's house W.

    Answer ...

  4. ((d))

    Find the area of the triangle WYZ.

    Answer ... km^2

  5. ((e))

    The smallest possible angle of depression of a point on the path WY from the top of the shopping mall Z is 25°.

    Find the height of shopping mall Z, giving your answer to the nearest metre.

    Answer ... m

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Worked solution

18((a))

1. Find the reverse bearing from $X$ to $Y$.

Bearing of XX from Y=064.1∘Y = 064.1^\circ.

Therefore, bearing of YY from XX is
064.1∘+180∘=244.1∘.064.1^\circ + 180^\circ = 244.1^\circ.

Answer

244.1∘244.1^\circ

18((e))

1. Identify the horizontal distance corresponding to the smallest angle of depression.

The smallest angle of depression occurs when the horizontal distance from ZZ to the point on path WYWY is greatest.

Since WZ=4.6 kmWZ=4.6\text{ km} and YZ=2.5 kmYZ=2.5\text{ km}, the furthest point on the path from ZZ is WW.

So the horizontal distance is
4.6 km.4.6\text{ km}.

2. Use the tangent ratio to find the height of the shopping mall.

Let the height of the shopping mall be hh km.

Using tan⁡θ=oppositeadjacent\tan \theta=\dfrac{\text{opposite}}{\text{adjacent}},
tan⁡25∘=h4.6\tan 25^\circ=\frac{h}{4.6}
h=4.6tan⁡25∘h=4.6\tan 25^\circ
h=2.145… kmh=2.145\ldots\text{ km}

Converting to metres,
2.145… km=2145… m2.145\ldots\text{ km}=2145\ldots\text{ m}

To the nearest metre,
h=2145 m.h=2145\text{ m}.

Answer

2145 m2145\text{ m}

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