Math, asked by NafishaIram, 1 year ago

the angles of elevation of the top of a tower from two points at distances of 4m and 9m from the base of the tower and in the same straight line with it are complementary. show that the height of the tower is 6m

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Answered by silentlover45
5

\underline\mathfrak{Given:-}

  • \: \: \: \: \: \: \: AB \: \: is \: \: the \: \: tower \: \: P, \: Q, \: \: are \: \: the \: \: period \: \: at \: \: distance \: \: {4cm} \: \: and \: \: {9cm}

  • \: \: \: \: \: \: \: PQ \: \: = \: \: {4cm}

  • \: \: \: \: \: \: \: QB \: \: = \: \: {9cm}

\underline\mathfrak{Prove \: That:-}

  • \: \: \: \: \: AB \: \: = \: \: {6cm}?

\underline\mathfrak{Solutions:-}

  • \: \: \: \: \: \: \: \alpha \: \: and \: \: \beta \: \: are \: \: supplementary

\: \: \: \: \: \fbox{\alpha \: + \: \beta \: \: = \: \: {90} \degree}

\: \: \: \: \: \therefore \: In \triangle \: ABP

\: \: \: \: \: \leadsto \: tan \alpha \: \: = \: \: \frac{AB}{BP} \: \: \: \: \: ...{(1)}.

\: \: \: \: \: \therefore \: In \triangle \: ABQ

\: \: \: \: \: \leadsto \: tan \beta \: \: = \: \: \frac{AB}{BQ}

\: \: \: \: \: \leadsto \: tan {({90} \: - \theta)} \: \: = \: \: \frac{AB}{BQ} \: \: \: \: \: {[Given \: \: \: \alpha \: + \: \beta \: \: = \: \: {90} \degree]}

\: \: \: \: \: \leadsto \: cot \: \alpha \: \: = \: \: \frac{AB}{BQ} \: \: \: \: \: {[tan \: {( {90} \: - \: \theta)} \: \: = \: \: {cot \: \theta}]}

\: \: \: \: \: \leadsto \: \frac{1}{tan \: \alpha} \: \: = \: \: \frac{AB}{BQ} \: \: \: \: \: \: {[cot \: \theta \: \: = \: \: \frac{1}{tan \: \theta}]}

\: \: \: \: \: \leadsto \: {tan \: \alpha} \: \: = \: \: \frac{BQ}{AB} \: \: \: \: \: ....{(2)}.

  • \: \: \: \: \: From \: \: Eq. {(1)} \: \: and \: \: Eq. \: {(2)}.

\: \: \: \: \: \leadsto \frac{AB}{BQ} \: \: = \: \: \frac{BQ}{AB}

\: \: \: \: \: \leadsto {AB} \: \times \: {AB} \: \: = \: \: {BQ} \: \times \: {BP}

  • \: \: \: \: \: put \: \: the \: \: value \: \: PB \: \: = \: \: {4m} \: \: and \: \: QB \: \: = \: \: {9m}

\: \: \: \: \: \leadsto {(AB)}^{2} \: \: = \: \: {4} \: \times \: {9}

\: \: \: \: \: \leadsto {(AB)}^{2} \: \: = \: \: {36}

\: \: \: \: \: \leadsto {AB} \: \: = \: \: \sqrt{36}

\: \: \: \: \: \leadsto {AB} \: \: = \: \: {6}

  • \: \: \: \: \: Hence, \: \: height \: \: of \: \: of \: \: the \: \: tower \: \: is \: \: {6m}.

\: \: \: \: \: \huge{Proved}.

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Answered by Anonymous
1

hope it is helpful for u ✔✔✔✔✔

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