The angle of elevation of the top of a tower from a point on the ground, which is $30m$ away from the foot of the tower, is ${30^0}$.Find the height of the tower.

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Hint: Try to visualise the problem with a diagram and use trigonometry.
Given in the problem the top of the tower when projected to a point makes an angle of ${30^0}$ with the ground. Also, the same point is $30m$away from the base of the tower.
Let $AC$be the tower and $B$be the point on ground.

Then $ABC$forms a right-angled triangle, right angled at $C$.

Then according to the problem $\angle ABC = {30^0}$ and $BC = 30m$.
Let $AC$,height of the tower be tower be denoted by $h$.

We need to find the height $h$ of the tower.
Looking at the right-angled triangle we need a relation between base and height.
We know that $\tan $trigonometric ratio gives the relation between base and height of the right-angled triangle,

$\tan \theta = \dfrac{{perpendicular}}{{base}}{\text{ (1)}}$
Here $perpendicular$is the height $h$and $base$ is the line segment $BC$.
Also $\theta = {30^0}$
Using the above in equation $(1)$,we get
  \tan \left( {{{30}^0}} \right) = \dfrac{h}{{BC}} \\
   \Rightarrow \tan \left( {{{30}^0}} \right) = \dfrac{h}{{30}} \\
We know that $\tan \left( {{{30}^0}} \right) = \dfrac{1}{{\sqrt 3 }}$
   \Rightarrow \dfrac{1}{{\sqrt 3 }} = \dfrac{h}{{30}} \\
   \Rightarrow \dfrac{{3 \times 10}}{{\sqrt 3 }} = h \\
   \Rightarrow h = 10\sqrt 3 m \cong 17.32m \\
Therefore, the height of the tower is $10\sqrt {3m} \cong 17.32m$.
Note: In the word problems related to trigonometry like above, try to draw the diagram first and then solve the question. Values of trigonometric ratios at basic angles must be kept in mind.