Art the foot of a mountain, the angle of elevation of its summit is 45∘. After ascending 1 km towards the mountain up an incline of 30∘, the elevation changes to 60∘ (as shown in the given figure). Fin dthe height of the mountain. [Given : √3=1.73.]
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(\frac{SD}{FD} = tan 45^\circ = 1 \Rightarrow SD = FD = h km\)
⇒∠FSD=∠SFD=45∘
∠ASC=180∘−(60∘+90∘)=30∘
⇒∠FSC=(45∘−30∘)=15∘ and ∠SFA=15∘
∴∠SAF=180∘−(15∘+15∘)=150∘
By the sine formula on DeltaSAF,we have
AFsin∠FSA=FSsin∠SAF⇒1sin15∘=√2hsin150∘=√2h30∘⇒2sqrt2(√3−1)=2√2h
[∴FS=√h2+h2=√2h and sin 15∘=(√3−1)2√2]
⇒h=1(√3)−1×(√3+1)(√3+1)=(√3+1)2=(1.73+1)2=2.732=1.365km.