From the top of the hill the angles of depression of two consecutive kilometer stones due east are found to be 45 and 30 respectively. find the height of the hill l
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The distance between two km. stone is 1000 mtrs.=AB
Let the hill makes an angle 45° with B point and with A point is 30°.
Let x be the height of the hill from the point O and let y be the distance from A to O.
∴tan 45°=x/(1000-y)
→1000-y=x
→y=1000-x ... (1)
again,
tan 30°=x/y
→1/√3=x/y
→y=x√3 ... (2)
Solving (1) and (2), we get
x=1000/(√3+1)=366.03 mtrs.
∴ y=1000√3/(√3+1)=633.97 mtrs.
Let the hill makes an angle 45° with B point and with A point is 30°.
Let x be the height of the hill from the point O and let y be the distance from A to O.
∴tan 45°=x/(1000-y)
→1000-y=x
→y=1000-x ... (1)
again,
tan 30°=x/y
→1/√3=x/y
→y=x√3 ... (2)
Solving (1) and (2), we get
x=1000/(√3+1)=366.03 mtrs.
∴ y=1000√3/(√3+1)=633.97 mtrs.
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