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प्रश्न
The angle of elevation of the top of a tower as observed form a point in a horizontal plane through the foot of the tower is 32°. When the observer moves towards the tower a distance of 100 m, he finds the angle of elevation of the top to be 63°. Find the height of the tower and the distance of the first position from the tower. [Take tan 32° = 0.6248 and tan 63° = 1.9626]
उत्तर
Let h be the height of the tower and the angle of elevation as observed from the foot of the tower is 32° and observed move towards the tower with distance 100 m then the angle of elevation becomes 63°.
Let BC = x and CD = 100
Now we have to find the height of the tower
So we use trigonometrical ratios.
In a triangle ABC,
`=> tan C = (AB)/(BC)`
`=> tan 63^@ = (AB)/(BC)`
`=> 1.9626 = h/x`
`=> x = h/1.9626`
Again in a triangle ABD
`=> tan D = (AB)/(BC + CD)`
`=> tan 32^@ = h/(x + 100)`
`=> 0.6248 = h/(x + 100)`
`=> x + 100 = h/0.6248`
`=> 100 = h/0.6248 - h/1.9626`
`=> 100 = (h xx 1.9626 - h xx 0.6248)/(0.6248 xx 1.9626)`
`=> 100 = (h(1.9626 - 0.6248))/(0.6248 xx 1.9626)`
`=> 100 = (h(1.3378))/(0.6248 xx 1.9626)`
=> 100 x 0.6248 x 1.9626 = h x 1.3378
`=> h = (100 xx 0.6248 xx 1.9626)/1.3378`
`=> 122.6232/1.3378`
=> 91.66
`=> x = 91.66/1.9626`
= 46.7
So distance of the first position from the tower is = 100 + 46.7 = 146.7 m
Hence the height of tower 91.66 m and the desires distance is 146.7 m
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