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In the figure, AB represents the tree, and AC represents the length of shadow casted.

Let, the height of tree be ${\text{h}}$ and length of shadow be ${\text{b}}$

So, ${\text{AB = h}}$and ${\text{AC = b}}$

Let, the angle of elevation of the sun be \[{\text{x}}\]

Then, in the figure, angle at C is ${\text{x}}$

Now, as given, the length of the shadow of a tree is $\sqrt 3 $ times its height,

So, ${\text{b = }}\sqrt {\text{3}} {\text{h}}$

Now, among various trigonometric ratios, those dealing with base and height are tangent and cotangent

We know that, for a right-angled triangle,

${\text{tan}\theta = }\dfrac{{{\text{height}}}}{{{\text{base}}}}$

Also, the triangle in the figure ABC is right angled at A

Base of the triangle is AC and height of the triangle is AB

So, applying ${\text{tan}\theta = }\dfrac{{{\text{height}}}}{{{\text{base}}}}$

Substituting, ${\theta = x}$

${\text{tanx = }}\dfrac{{{\text{AB}}}}{{{\text{AC}}}}$

${\text{tanx = }}\dfrac{{\text{h}}}{{\text{b}}}$

Now, substituting, ${\text{b = }}\sqrt {\text{3}} {\text{h}}$

${\text{tanx = }}\dfrac{{\text{h}}}{{\sqrt {\text{3}} {\text{h}}}}$

${\text{tanx = }}\dfrac{{\text{1}}}{{\sqrt {\text{3}} }}$

Now, we know that,

${\text{tan}}\left( {{\text{30}^\circ }} \right){\text{ = }}\dfrac{{\text{1}}}{{\sqrt {\text{3}} }}$

Therefore, ${\text{x = 30}^\circ }$

Hence, the angle of elevation of the sun is ${\text{30}^\circ}$

\[

{\text{sin}\theta = }\dfrac{{{\text{height}}}}{{{\text{hypotenuse}}}} \\

{\text{cos}\theta = }\dfrac{{{\text{base}}}}{{{\text{hypotenuse}}}} \\

{\text{tan}\theta = }\dfrac{{{\text{height}}}}{{{\text{base}}}} \\

{\text{sec}\theta = }\dfrac{{{\text{hypotenuse}}}}{{{\text{base}}}} \\

{\text{cosec}\theta = }\dfrac{{{\text{hypotenuse}}}}{{{\text{height}}}} \\

{\text{cot}\theta = }\dfrac{{{\text{base}}}}{{{\text{height}}}} \\

\]

Students must remember them to solve these types of questions.