Question

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Answer 1

Question:

A vertical tower is surmounted by a flag staff of height 5m. At a poiyon the ground the angles of elevation of the bottom and the top of flag staff are 45° and 60° respectively. Find the height of the tower .

Answer:

6.825 m

Solution:

First off all , let's plot a rough diagram to describe the given situation.

Let OC be the tower with its foot at the point O.

Let BC be the flag staff with its top at the point B and the bottom at the point C.

Also, the angle of elevation of the point C and B from the point A on the ground are 45° and 60° respectively.

{ For figure, please refer to the attachment }

Now,

In ∆AOC ,

=> tan45° = OC/OA

=> 1 = OC/OA

=> OA = OC

=> OA = OC = x m (say) -------(1)

Also,

In ∆OAB ,

=> tan60° = OB/OA

=> √3 = (OC + BC)/OA

=> √3 = (x + 5)/x. {using eq-(1)}

=> √3x = x + 5

=> √3x - x = 5

=> x(√3 - 1) = 5

=> x = 5/(√3-1)

=> x = 5•(√3+1)/(√3-1)•(√3+1)

{rationalising the denominator}

=> x = 5•(√3+1)/(3-1)

=> x = 5•(√3+1)/2

=> x = 5•(1.73+1)/2

=> x = 5•2.73/2

=> x = 13.65/2

=> x = 6.825 m

Hence,

The height of the tower OC is x m , ie; 6.825 m .

Answer 2

$\begin{lgathered} \red{\bf{Given}}\begin{cases} \sf \: flag \: staff \: height = 5m. \\ \sf \: angle \: of \: elevation \: from \: bottom = 45 \\ \sf \: angle \: of \: elevation \: from \: top = 60 \\ \bf \: \green{tower \: height \: =?} \: \end{cases}\end{lgathered}$

$\Large\bold\star\underline{\underline\textbf{Formula\:used}}$

• Tan @ = Perpendicular/Base
• Tan 45° = 1
• Tan 60° = √3
• √3 = 1.73

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$\large\star{\underline{\tt{\red{Answer}}}}\star$

$\red{\textbf{Refer To image First}}:$

From image we can see that,

→ BC = Base of Triangle ∆BDC and ∆ABC both.

→ Perpendicular of ∆BDC = BD = x metre.

→ Perpendicular of ∆ABC = AB = BD + AD = (x + 5) metre .

→ Angle BCD = 45°

→ Angle ACB = 60°

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$\bf \: in \: \red\triangle BDC , \: we \: have : \\ \\ \red\longrightarrow \: tan45 = \frac{BD}{BC} \\ \\ \red\longrightarrow \: 1 = \frac{x}{BC} \\ \\ \red\longrightarrow \: \orange{\boxed{ \bf \: BC \: = \: x}} \: - - - \sf \: equation(1)$

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$\bf \: Now, in \: \red\triangle ABC, we \: have : \\ \\ \red\leadsto \: tan60 = \frac{AB}{BC} \\ \\ \: \red\leadsto \: \sqrt{3} = \frac{x + 5}{BC} \\ \\ \red\leadsto \: BC \: = \frac{x + 5}{ \sqrt{3} } \\ \\ \sf \: rationalizing \: the \: denominator \: now \\ \\ \red\leadsto \: BC \: = \frac{x + 5}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } \\ \\ \red\leadsto \: \purple{\boxed{ \bf \: BC = \frac{ \sqrt{3}( x + 5) }{3}}} - - - \sf \: equation(2)$

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Now, since Equation (1) = Equation(2) = BC ,

Comparing both Now, we get,,

$\red{\boxed\implies} \: x = \dfrac{ \sqrt{3}(x + 5) }{3} \\ \\ \red{\boxed\implies} \: 3x = \sqrt{3}x + 5 \sqrt{3} \\ \\ \red{\boxed\implies} \: 3x - \sqrt{3} x = 5 \sqrt{3} \\ \\ \red{\boxed\implies} \: x(3 - \sqrt{3} ) = 5 \sqrt{3} \\ \\ \red{\boxed\implies} \: x = \frac{5 \sqrt{3} }{(3 - \sqrt{3}) } \\ \\ \blue{\sf \: putting \: \sqrt{3} = 1.73 \: now} \\ \\ \red{\boxed\implies} \: x = \frac{5 \times 1.73}{3 - 1.73} \\ \\ \red{\boxed\implies} \: x = \frac{5 \times 1.73}{1.27} \\ \\ \red{\boxed\implies} \: \: \pink{ \large \boxed{ \bf \: x = 6.811m}}$

Hence, Height of Tower will be 6.8 m (Approx) .

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