W, X, Y and Z are points on a horizontal ground and PW is a vertical flag pole. WX=7 mWX = 7\ \text{m}, WZ=3 mWZ = 3\ \text{m}, XY=10 mXY = 10\ \text{m}, ∠WYZ=20∘\angle WYZ = 20^\circ and ∠WZY=28.6∘\angle WZY = 28.6^\circ.

  1. (a)

    Calculate WY.

  2. (b)

    Show that ∠WXY=20.2∘\angle WXY = 20.2^\circ, correct to 1 decimal place.

  3. (c)

    The bearing of Z from Y is 308°. Find the bearing of W from Z.

  4. (d)

    Given that PX = 8 m, calculate the height of the flag pole PW.

  5. (e)

    T is a point along XY. Find the greatest angle of elevation of the top of the flag pole P from T.

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Worked solution

4(a)

1. Find the third angle in triangle WZY.

In △WZY\triangle WZY,
∠ZWY=180∘−20∘−28.6∘=131.4∘.\angle ZWY = 180^\circ - 20^\circ - 28.6^\circ = 131.4^\circ.

2. Apply the sine rule to find WY.

By the sine rule,
WYsin⁡28.6∘=WZsin⁡20∘.\frac{WY}{\sin 28.6^\circ}=\frac{WZ}{\sin 20^\circ}.
Since WZ=3WZ=3,
WY=3sin⁡28.6∘sin⁡20∘=4.197…WY=\frac{3\sin 28.6^\circ}{\sin 20^\circ}=4.197\ldots
WY≈4.20 m.WY\approx 4.20\text{ m}.

Answer

WY=4.68 mWY = 4.68\text{ m}

4(c)

1. Find the bearing of Y from Z using the reverse bearing.

The bearing of ZZ from YY is 308∘308^\circ.

Therefore the bearing of YY from ZZ is
308∘−180∘=128∘.308^\circ - 180^\circ = 128^\circ.

2. Use the angle at Z to find the bearing of W from Z.

At ZZ,
∠WZY=28.6∘.\angle WZY = 28.6^\circ.

From the diagram, ZWZW is clockwise from ZYZY, so
bearing of W from Z=128∘+28.6∘=156.6∘.\text{bearing of }W\text{ from }Z = 128^\circ + 28.6^\circ = 156.6^\circ.

To the nearest degree, this is
157∘.157^\circ.

Answer

157∘157^\circ

4(d)

1. Use Pythagoras' theorem to find the vertical height PW.

Since PWPW is vertical and WXWX is on horizontal ground, ∠PWX=90∘\angle PWX = 90^\circ.

In right-angled triangle PWXPWX,
PX2=PW2+WX2PX^2 = PW^2 + WX^2
82=PW2+728^2 = PW^2 + 7^2
PW2=64−49=15PW^2 = 64 - 49 = 15
PW=15=3.87 m correct to 3 s.f.PW = \sqrt{15} = 3.87\text{ m} \text{ correct to 3 s.f.}

Answer

PW=3.87 mPW = 3.87\text{ m}

4(e)

1. Identify the position of T that gives the greatest angle of elevation.

Since PWPW is fixed, the angle of elevation θ\theta from TT satisfies
tan⁡θ=PWTW.\tan \theta=\frac{PW}{TW}.
The angle θ\theta is greatest when TWTW is as small as possible.

The shortest distance from WW to the line segment XYXY is the perpendicular distance from WW to XYXY.

2. Find the shortest possible horizontal distance from W to T on XY.

Using ∠WXY=20.2∘\angle WXY=20.2^\circ from part (b),
shortest WT=7sin⁡20.2∘.\text{shortest }WT=7\sin 20.2^\circ.
WT=2.41 m(approximately)WT=2.41\text{ m} \quad \text{(approximately)}

3. Calculate the greatest angle of elevation.

Using PW=3.87 mPW=3.87\text{ m} from part (d),
tan⁡θ=3.872.41.\tan \theta=\frac{3.87}{2.41}.
θ=tan⁡−1(3.872.41)\theta=\tan^{-1}\left(\frac{3.87}{2.41}\right)
θ=58.0∘(to 1 decimal place)\theta=58.0^\circ \quad \text{(to 1 decimal place)}

Answer

58.0∘58.0^\circ

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