Question

Art the foot of a mountain, the angle of elevation of its summit is ${45}^{\circ }$. After ascending 1 km towards the mountain up an incline of ${30}^{\circ }$, the elevation changes to ${60}^{\circ }$ (as shown in the given figure). Fin dthe height of the mountain. [Given : $\sqrt{3}=\mathrm{1.73.}]$

Answer 1

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(\frac{SD}{FD} = tan 45^\circ = 1 \Rightarrow SD = FD = h km\)

$\Rightarrow \angle FSD=\angle SFD={45}^{\circ }$

$\angle ASC={180}^{\circ }-({60}^{\circ }+{90}^{\circ })={30}^{\circ }$

$\Rightarrow \angle FSC=({45}^{\circ }-{30}^{\circ })={15}^{\circ }and\angle SFA={15}^{\circ }$

$\therefore \angle SAF={180}^{\circ }-({15}^{\circ }+{15}^{\circ })={150}^{\circ }$

By the sine formula on $DeltaSAF,$we have

$\frac{AF}{sin\angle FSA}=\frac{FS}{sin\angle SAF}\Rightarrow \frac{1}{sin{15}^{\circ }}=\frac{\sqrt{2}h}{sin{150}^{\circ }}=\frac{\sqrt{2}h}{{30}^{\circ }}\Rightarrow \frac{2sqrt2}{(\sqrt{3}-1)}=2\sqrt{2}h$

$[\therefore FS=\sqrt{h^2+h^2}=\sqrt{2}handsin{15}^{\circ }=\frac{(\sqrt{3}-1)}{2\sqrt{2}}]$

$\Rightarrow h=\frac{1}{\left(\sqrt{3}\right)-1}\times \frac{(\sqrt{3}+1)}{(\sqrt{3}+1)}=\frac{(\sqrt{3}+1)}{2}=\frac{(1.73+1)}{2}=\frac{2.73}{2}=1.365km.$