Question

A pole 3m high casts a shadow √3 m long on the ground, then find the sun’s
elevation.

Answer 1

Given :-

• Height of pole = 3 m
• Length of shadow = √3 m

$\:$

To find :-

Angle of sun's elevation = θ = ?

$\:$

Solution :-

To solve such questions, we apply the concept of trigonometry.

$\to \: \sf{tan \theta = \dfrac{ \sqrt{3} }{3}}$

$\to \: \sf{tan \theta = \dfrac{ \cancel{\sqrt{3}} }{ \sqrt{3} \times \cancel{\sqrt{3}} } }$

$\to \: \sf{tan \theta = \dfrac{1}{ \sqrt{3} } }$

$\to \: \sf{ \theta = tan {}^{ - 1} \bigg(\dfrac{1}{ \sqrt{3} } \bigg)}$

$\to \: \boxed{ \bf{\theta = {30}^{ \circ} }}$

∴ The angle of elevation of Sun = 30°

Know more:

tan theta is the opposite side divided by the adjacent side.

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

Answer 2

Gɪᴠᴇɴ Iɴғᴏ :-

• A pole 3m high casts a shadow √3 m long on the ground.

Tᴏ Fɪɴᴅ :-

• Sun’s elevation.

Cᴏɴᴄᴇᴘᴛ Usᴇᴅ :-

• Here, the concept of trigonometry is used.

$\sf \quad \bull \: \tan \theta \: \: ➪ \: \: \dfrac{length \: of \: the \: shadow}{length \: of \: the \: pole}$

Rᴇǫᴜɪʀᴇᴅ Sᴏʟᴜᴛɪᴏɴ :-

Lᴇᴛ, (θ) be the elevation of the Sun.

$\\ \sf ➠ \: \tan \theta \: \: ➪ \: \: \dfrac{length \: of \: the \: shadow}{length \: of \: the \: pole}$

$\tt ➠ \: \tan \theta \: \: ➪ \: \: \dfrac{ \sqrt{3} }{3 }$

$\sf ➠ \: \tan \theta \: \: ➪ \: \: \dfrac {\cancel{ \sqrt{3} } \: {}^{1} }{ \sqrt{3} \times \cancel{\sqrt{3} } \: _{1} }$

Hᴇʀᴇ , √3 and √3 is cancelled.

$\\ \sf ➠ \: \tan \theta \: \: ➪ \: \: \dfrac {{1} }{ \sqrt{3} }$

$\sf ➠ \: \theta \: \: ➪ \: \: \tan {}^{ - 1} \: \: \bigg(\dfrac {1 }{ \sqrt{3} }\bigg)$

Hᴇʀᴇ, ᴡᴇ ᴀʀᴇ ᴛᴀᴋɪɴɢ tan ᴛᴏ ᴛʜᴇ RHS ᴀɴᴅ ᴡᴇ ʜᴀᴠᴇ ᴍᴀᴋᴇᴅ tan ➠ tan ^(-1).

$: \implies \: \underline{ \boxed{ \bf{ \theta \: \: ➠ \: \: 30 {}^{ \circ} }}}$

$\bf{\therefore{ \underline{Hence, \: {\sf{angle \: of \: elevation \: of \: sun \: is \: }}30 {}^{ \circ} .}}}$

✯ Hope it helps u ✯