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Step-by-step explanation:
AnswEr :
⋆ Refrence of Image is in the Diagram :
• In ∆ YRQ, we have :
\begin{lgathered}\implies\tt\tan(45 \degree) = \dfrac{QR}{YR} \\ \\\implies \tt1 = \dfrac{x}{YR} \\ \\\implies \tt YR = x \\ \\\implies \tt XP = x \qquad \qquad [ \because YR = XP]\end{lgathered}
⟹tan(45°)=
YR
QR
⟹1=
YR
x
⟹YR=x
⟹XP=x[∵YR=XP]
• In ∆ XPQ, we have :
\begin{lgathered}\implies\tt\tan(60 \degree) = \dfrac{PQ}{PX} \\ \\\implies\tt \sqrt{3} =\dfrac{(x + 40)}{x} \\ \\\implies\tt \sqrt{3}x = x + 40 \\ \\\implies\tt \sqrt{3}x - x =40 \\ \\\implies\tt x( \sqrt{3} - 1) = 40 \\ \\\implies\tt x = \dfrac{40}{( \sqrt{3} - 1)} \\\\\implies\tt x = \dfrac{40}{( \sqrt{3} - 1) } \times\dfrac{( \sqrt{3} + 1)}{(\sqrt{3} + 1)} \\ \\\implies\tt x = \dfrac{40( \sqrt{3} + 1)}{ {( \sqrt{3} )}^{2} - {(1)}^{2} } \\ \\\implies\tt x = \dfrac{40( \sqrt{3} + 1)}{3 - 1} \\ \\\implies\tt x = \dfrac{ \cancel40(\sqrt{3} + 1)}{\cancel2} \\ \\\implies\tt x =20( \sqrt{3} + 1) \\ \\\implies\tt x =20(1.732 + 1) \qquad \scriptsize[\sqrt{3} = 1.732]\\ \\\implies\tt x =20 \times 2.732 \\\\\implies \boxed{\pink{\tt x =54.64 \:metres}}\end{lgathered}
⟹tan(60°)=
PX
PQ
⟹
3
=
x
(x+40)
⟹
3
x=x+40
⟹
3
x−x=40
⟹x(
3
−1)=40
⟹x=
(
3
−1)
40
⟹x=
(
3
−1)
40
×
(
3
+1)
(
3
+1)
⟹x=
(
3
)
2
−(1)
2
40(
3
+1)
⟹x=
3−1
40(
3
+1)
⟹x=
2
4
0(
3
+1)
⟹x=20(
3
+1)
⟹x=20(1.732+1)[
3
=1.732]
⟹x=20×2.732
⟹
x=54.64metres
• H E I G H T⠀O F⠀T O W E R :
\begin{lgathered}\longrightarrow \tt Height \:of \:Tower = PQ\\\\\longrightarrow \tt Height \:of \:Tower = (QR+PR)\\\\\longrightarrow \tt Height \:of \:Tower = (x+40\:metres)\\\\\longrightarrow \tt Height \:of \:Tower = (54.64 + 40)\:metres\\\\\longrightarrow\large\boxed{\red{\tt Height \:of \:Tower = 94.64\:metres}}\end{lgathered}
⟶HeightofTower=PQ
⟶HeightofTower=(QR+PR)
⟶HeightofTower=(x+40metres)
⟶HeightofTower=(54.64+40)metres
⟶
HeightofTower=94.64metres
