The length of a string between a kite and a point on the ground is 85 m- If the string makes an angle 0 with level ground such that tan - - 15-8- how high is the kite -

The correct option is B 75 mGiven that : AC = 90 m and tan θ = 158 = BCAB = hAB h = AB tan θ....(1) Now, In ABC, (AC)^2 = (AB)^ ...

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Standard X

Mathematics

Calculating Heights and Distances

The length of...

Question

The length of a string between a kite and a point on the ground is $85 m$ . If the string makes an angle $θ$ with level ground such that $tan θ = \frac{15}{8}$ , how high is the kite ?

75

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78.05

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226

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70

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Solution

The correct option is B. $75$ m
Given that : $A C = 90$ m and
$tan θ = \frac{15}{8} = \frac{B C}{A B}$
Let $B C = 15 x$ and $A B = 8 x$
By Pythagoras Theorem,
$A C^{2} = B C^{2} + A B^{2}$
$85^{2} = (15 x)^{2} + (8 x)^{2}$
$7225 = 225 x^{2} + 64 x^{2}$
$7225 = 289 x^{2}$
$25 = x^{2}$
$∴ x = 5$
Hence, $B C = 15 \times 5 = 75 m$

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The length of a string between a kite and a point on the ground is 85m. If the string makes an angle θ with level ground such that tanθ=158, how high is the kite ?

The correct option is B.75 mGiven that : AC=90m and tanθ=158=BCABLet BC=15x and AB=8xBy Pythagoras Theorem,AC2=BC2+AB2852=(15x)2+(8x)27225=225x2+64x27225=289x225=x2∴x=5Hence, BC=15×5=75m

The length of a string between a kite and a point on the ground is $85 m$ . If the string makes an angle $θ$ with level ground such that $tan θ = \frac{15}{8}$ , how high is the kite ?