Question

At a point on level ground the angle of elevation of a vertical tower to be found such that its tangent is 5/12. On walking 192 m towards the tower, the tangent of the angle is found to be 3/4. Find the height if the tower.

Plz help me to solve this one ASAP!

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Answer 1

: Required Answer

➝ Let AB be the vertical tower and C and D be two points Such that CD = 192 m.

➝ Let ∠ACB = θ and ∠ADB = a

$\implies$ $\small\sf \pink{given}$

$\implies$$\tan = \frac{5}{12}$

$\implies$ $\frac{ab}{cd } = \frac{5}{12}$

$\implies$ $ab = \frac{5}{12} \: bc \: (i)$

$\implies$ also $\tan \: a \: = \frac{3}{4}$

$\implies$ $\frac{ab}{bd} = \frac{3}{4}$

$\implies$ $\frac{5}{12} \: bc$$\implies$ $\frac{12}{bd} = \: \frac{3}{4}$

$\implies$ $\frac{192 + bd}{bd} = \frac{3}{4} </p><p></p><p> \times \frac{12}{5}$

$\implies$ $bd = 240 \: m$

$\implies$ $bc \:=(192 + 240) = 432 \: m$

$\implies$ By (i) AB

$= \frac{5}{12} \times 432 = 180 \: m$

$\small\sf \red{hence \: the \: hight \: of \: the \: tower \: is \: 180 \: m}$

Answer 2

Step-by-step explanation:

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