8 Fig. 8 shows a searchlight, mounted at a point A, 5 metres above level ground. Its beam is in the shape of a cone with axis AC , where C is on the ground. AC is angled at \(\alpha\) to the vertical. The beam produces an oval-shaped area of light on the ground, of length DE . The width of the oval at C is GF . Angles DAC, EAC, FAC and GAC are all \(\beta\).
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\caption{Fig. 8}
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In the following, all lengths are in metres.
- Find AC in terms of \(\alpha\), and hence show that \(\mathrm { GF } = 10 \sec \alpha \tan \beta\).
- Show that \(\mathrm { CE } = 5 ( \tan ( \alpha + \beta ) - \tan \alpha )\).
$$\text { Hence show that } \mathrm { CE } = \frac { 5 \tan \beta \sec ^ { 2 } \alpha } { 1 - \tan \alpha \tan \beta } \text {. }$$
Similarly, it can be shown that \(\mathrm { CD } = \frac { 5 \tan \beta \sec ^ { 2 } \alpha } { 1 + \tan \alpha \tan \beta }\). [You are not required to derive this result.]
You are now given that \(\alpha = 45 ^ { \circ }\) and that \(\tan \beta = t\). - Find CE and CD in terms of \(t\). Hence show that \(\mathrm { DE } = \frac { 20 t } { 1 - t ^ { 2 } }\).
- Show that \(\mathrm { GF } = 10 \sqrt { 2 } t\).
For a certain value of \(\beta , \mathrm { DE } = 2 \mathrm { GF }\).
- Show that \(t ^ { 2 } = 1 - \frac { 1 } { \sqrt { 2 } }\).
Hence find this value of \(\beta\).