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Question Number 12962 by tawa last updated on 07/May/17

Answered by sandy_suhendra last updated on 09/May/17

Commented by sandy_suhendra last updated on 09/May/17

$$\mathrm{a})\mathrm{from}\mathrm{picture}\left(\mathrm{i}\right)$$$$\mathrm{T}=\mathrm{top}\mathrm{of}\mathrm{the}\mathrm{flagstaff}$$$$\mathrm{PR}=\sqrt{{100}^2+{70}^2}=122.07\mathrm{m}$$$$\tan 15°=\frac{\mathrm{TP}}{\mathrm{PR}}\Rightarrow \mathrm{TP}=\mathrm{PR}\tan 15°$$$$=122.07\times 0.268=32.7\mathrm{m}$$$$\mathrm{b})\mathrm{from}\mathrm{picture}\left(\mathrm{ii}\right)$$$$\mathrm{MP}=\frac{1}{2}\mathrm{PQ}=50\mathrm{m}$$$$\tan \alpha =\frac{\mathrm{TP}}{\mathrm{MP}}=\frac{32.7}{50}=0.654$$$$\alpha =33.2°$$

Commented by tawa tawa last updated on 10/May/17

$$\mathrm{wow},\mathrm{G}\mathit{o}\mathit{d}\mathit{b}\mathit{l}\mathit{e}\mathit{s}\mathit{s}\mathit{y}\mathit{o}\mathit{u}\mathit{s}\mathit{i}\mathit{r}.$$

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