Question

In the adjoining figure , a tower AB is 30 m high and BC its shadow on the ground is 20root3 long . Find the sun's altitude

Answer 1

Correct Question :

A tower AB is 20 m high and BC , its shadow on ground, is 20√3 m long. Find the sun's altitude.

AnswEr :

⋆ Refrence of Image is in the Diagram :

$\setlength{\unitlength}{1.3 pt}\begin{picture}(100,100)(0,0)\put(0,10){\line(1,0){95}}\put(40,10){\line(0,1){55}}\put(95,10){\line(-1,1){85}}\multiput(0,65)(6,0){7}{\line(1,0){3}}\qbezier(90,15)(80,20)(82,10)\put(10,95){\circle*{20}}\put(24,40){ \sf{20\:m} }\put(15,50){ \bf{Tower} }\put(60,0){ \sf{20\sqrt{3}\:m} }\put(10,105){ \bf Sun }\put(40,3){ \sf B }\put(40,68){ \sf A }\put(95,3){ \sf C }\end{picture}$

• Here :

• AB (Tower) = 20 metre
• BC (Shadow) = 20√3 metre
• ∠ ACB (Sun's Altitude) = ?

$\rule{150}{1}$

$\underline{\bigstar\:\textsf{In Triangle ABC, Using Trigonometric Value :}}$

$\dashrightarrow\:\:\tt \tan(\theta) =\dfrac{Perpendicular}{Base}\\\\\\\dashrightarrow\:\:\tt \tan(ACB)=\dfrac{AB}{BC}\\\\\\\dashrightarrow\:\:\tt \tan(ACB)=\dfrac{20}{20\sqrt{3}}\\\\\\\dashrightarrow\:\:\tt \tan(ACB)= \dfrac{1}{\sqrt{3}}\\\\{\scriptsize\qquad\bf{\dag}\:\:\sf{\tan(30\degree)=\dfrac{1}{\sqrt{3}}}}\\\\\dashrightarrow\:\:\tt \tan(ACB)=\tan(30\degree)\\\\\\\dashrightarrow\:\: \underline{\boxed{\tt \angle\:ACB = 30 \degree}}$

⠀

$\therefore\:\underline{\textsf{Hence, Sun's Altitude is at \textbf{30\degree}}}.$

$\rule{200}{2}$

$\bigstar\:\sf Trigonometric\:Values :\\\boxed{\begin{tabular}{c|c|c|c|c|c}Radians/Angle & 0 & 30 & 45 & 60 & 90\\\cline{1-6}Sin \theta & 0 & \dfrac{1}{2} & \dfrac{1}{\sqrt{2}} & \dfrac{\sqrt{3}}{2} & 1\\\cline{1-6}Cos \theta & 1 & \dfrac{\sqrt{3}}{2}& \dfrac{1}{\sqrt{2}}& \dfrac{1}{2}&0\\\cline{1-6}Tan \theta&0& \dfrac{1}{\sqrt{3}}&1&\sqrt{3}&Not D \hat{e} fined\end{tabular}}$

Answer 2

Answer:

• Height of the tower = AB = 20 m

• Length of the shadow = BC = 20√3 m

• The Altitude of the Sun = < ACB

⇒ tan <ACB = AB/CB

= 20/20√3

= 1/√3

⇒ tan <ACB = tan 30°

<ACB = 30°