Question

if the altitude of the sun is at 60 degree then the height of the vertical tower that will cast a shadow of length 20 meters is

Answer 1

Given:-

• Angle of elevation = 60°
• Length of Shadow = 20m

Find:-

• Length of Tower.

Diagram:-

Let, AB be the tower whose height is 'h'm and BC be the shadow of 20m.

$\setlength{\unitlength}{1cm}\begin{picture}(6,5)\linethickness{.4mm}\put(1,1){\line(1,0){4.5}}\put(1,1){\line(0,1){3.5}}\qbezier(1,4.5)(1,4.5)(5.5,1)\put(1.02,1.02){\framebox(0.2,0.3)}\put(.7,4.8){\large\bf A}\put(.8,.3){\large\bf B}\put(5.8,.3){\large\bf C}\qbezier(4.4,1)(4.3,1.25)(4.6,1.7) \put(3.8,1.3){ \bf 60^{\circ} }\put(2.8,0.7){\bf 20m}\put(0.1,3){ \bf 'h'm } \end{picture}$

Solution:-

In ∆ABC

we, know

$\implies\sf \dfrac{P}{B} = \tan C$

$\implies\sf \dfrac{P}{B} = \tan {60}^{ \circ}$

$\sf where \small{\begin{cases} \sf P = \text{'h'm}\\ \sf B = 20m\\ \sf tan \: 60^{\circ} = \sqrt{3} \end{cases}}$

Substituting these values:-

$\dashrightarrow\sf \dfrac{P}{B} = \tan {60}^{ \circ}$

$\dashrightarrow\sf \dfrac{h}{20} = \sqrt{3}$

Cross-multiplication:-

$\dashrightarrow\sf \dfrac{h}{20} = \dfrac{\sqrt{3}}{1}$

$\dashrightarrow\sf h = 20 \sqrt{3}m$

$\underline{\boxed{\sf\therefore Height\:of\:the\:tower\:is\:20\sqrt{3}m}}$

$\setlength{\unitlength}{1cm}\begin{picture}(6,5)\linethickness{.4mm}\put(1,1){\line(1,0){4.5}}\put(1,1){\line(0,1){3.5}}\qbezier(1,4.5)(1,4.5)(5.5,1)\put(1.02,1.02){\framebox(0.2,0.3)}\put(.7,4.8){\large\bf A}\put(.8,.3){\large\bf B}\put(5.8,.3){\large\bf C}\qbezier(4.4,1)(4.3,1.25)(4.6,1.7) \put(3.8,1.3){ \bf 60^{\circ} }\put(2.8,0.7){\bf 20m}\put( - 0.4,3) { \bf 20 \sqrt{3}m } \end{picture}$

Answer 2

Answer:

h=20√3.

Step-by-step explanation:

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