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Question
The length of a string between a kite and a point on the ground is 90 meters. If the string makes an angle O with the ground level such that tan O = 15/8, how high is the kite? Assume that there is no slack in the string.
Solution
Length of string between the point on ground and kite = 90
Angle made by the string with the ground is θ and `tan theta = 15/8`
`=> theta = tan^(-1) (15/8)`
The height of the kite be Hm
If we represent the above data in the figure as shown then it forms right-angled triangle ABC.
We have,
`tan theta = (AB)/(BC)`
`=> 15/8 = H/(BC)`
`=> BC = 84/15` .......(1)
In ΔABC , by Pythagoras theorem we have
`AC^2 = BC^2 + AB^2`
`=> 90^2 = ((8H)/(15))^3 + H^2`
`=> 90^2 = ((8H)^2 + (15H)^2)/15^2`
`=> H^2 (8^2 + 15^2) = 90^2 xx 15^2`
`=> H^2 (64 + 225) = (90 xx 15)^2`
`=> H^2 = (90 xx 15)^2/289`
`= H^2= ((90 xx 15)/17)^2`
`=> H = (90 xx 15)/17 = 79.41`
∴ height of kite from ground H = 79.41 m
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