question_answer

At the foot of mountain, the elevation of its summit is \[45{}^\circ .\] After ascending 2 km towards the mountain upon an incline of \[30{}^\circ ,\]the elevation changes to\[60{}^\circ \]. The height of the mountain is

A) \[(\sqrt{3}-1)\,km\]

B) \[(\sqrt{3}+1)\,km\]

C) \[(\sqrt{3}-2)\,km\]

D) \[(\sqrt{3}+2)\,km\]

Correct Answer: B

Solution :

[b] Let       \[AB=h\,km\] \[\therefore \]In \[\Delta {\mathrm O}\Alpha \Beta ,\]\[\tan 45{}^\circ =\frac{AB}{OB};\]\[OB=h\,km\] In \[\Delta {\mathrm O}L\Mu ,\]\[OM=2\cos 30{}^\circ =\sqrt{3}\,km\] \[\therefore \]      \[LN=BM=(h-\sqrt{3})\,KM\] In \[\Delta \,OLM,\]\[\sin 30{}^\circ =\frac{AN}{LM}\] \[LM=2\sin 30{}^\circ =1\,km\] \[\therefore \]      \[BN+LM+1km\] In \[\Delta \Alpha LN,\]\[\tan 60{}^\circ =\frac{AN}{LN}\] \[\sqrt{3}=\frac{AB=BN}{LN};\]\[\sqrt{3}=\frac{h-1}{h-\sqrt{3}}\]             \[\sqrt{3}h-3=h-1\]                         \[=h\frac{2}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3+1}}\]                         \[h=(\sqrt{3}+1)\,km\]