A tree stands vertically on a hill side which makes an angle of $15^{\circ}$ with the horizontal. From a point on the ground $35$ m down the from the base of the tree, the angle of elevation of the tree is $60^{\circ}$ . Find the height of the tree.

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Solution

In $△ O B T$
$\frac{h}{sin 45^{o}} = \frac{35}{sin 30^{o}}$
$\Rightarrow h = \frac{35. (\frac{1}{\sqrt{2}})}{\frac{1}{2}}$
$= 35 \sqrt{2} m$

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A tree stands vertically on a hill side which makes an angle of 15∘ with the horizontal. From a point on the ground 35 m down the from the base of the tree, the angle of elevation of the tree is 60∘. Find the height of the tree.

In △OBThsin45o=35sin30o⇒h=35.(1√2)12=35√2 m

A tree stands vertically on a hill side which makes an angle of $15^{\circ}$ with the horizontal. From a point on the ground $35$ m down the from the base of the tree, the angle of elevation of the tree is $60^{\circ}$ . Find the height of the tree.

In $△ O B T$
$\frac{h}{sin 45^{o}} = \frac{35}{sin 30^{o}}$
$\Rightarrow h = \frac{35. (\frac{1}{\sqrt{2}})}{\frac{1}{2}}$
$= 35 \sqrt{2} m$