
The horizontal distance between two trees is 60 m. The angle of depression of the top of the first tree as seen from the top of the second tree is $45{}^\circ $ . If the height of the second tree is 80 m, find the height of the first tree.
Answer
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Hint: Assume the height of the first tree be ‘h’. Draw a rough diagram of the given conditions and then use the formula $\tan \theta =\dfrac{\text{perpendicular}}{\text{base}}$ in the different right angle triangles and substitute the given values to get the height.
Complete step-by-step answer:
Let us start with the question by drawing a representative diagram of the situation given in the question.
According to the above figure:
BD is the first tree and we have assumed its height to be ‘h’ meters. AE is the second tree with height 80 meters as given in the question.
Now, in right angle triangle ABM,
$\angle ABM=45{}^\circ $ , as it is equal to the angle of depression mentioned in the question and shown in the diagram according to the property of alternate interior angles.
We know that, $\tan \theta =\dfrac{\text{perpendicular}}{\text{base}}$. Therefore,
$\tan {{45}^{\circ }}=\dfrac{AM}{BM}$
Since, BM = DE = 60 m, because they are opposite sides of the rectangle BDEM. Therefore,
$\tan {{45}^{\circ }}=\dfrac{AM}{60}$
Substituting $\tan {{45}^{\circ }}=1$, we get,
$AM=60$
Now from the figure, we can say that $BD=80-60=20$
So, the height of the first tree is 20 m.
Note: Be careful about the trigonometric ratios which you are using for the question. We have used the tangent of the given angle because we have to find the height and we have a common base in two right angle triangles that can be easily cancelled.
Complete step-by-step answer:
Let us start with the question by drawing a representative diagram of the situation given in the question.
According to the above figure:
BD is the first tree and we have assumed its height to be ‘h’ meters. AE is the second tree with height 80 meters as given in the question.
Now, in right angle triangle ABM,
$\angle ABM=45{}^\circ $ , as it is equal to the angle of depression mentioned in the question and shown in the diagram according to the property of alternate interior angles.
We know that, $\tan \theta =\dfrac{\text{perpendicular}}{\text{base}}$. Therefore,
$\tan {{45}^{\circ }}=\dfrac{AM}{BM}$
Since, BM = DE = 60 m, because they are opposite sides of the rectangle BDEM. Therefore,
$\tan {{45}^{\circ }}=\dfrac{AM}{60}$
Substituting $\tan {{45}^{\circ }}=1$, we get,
$AM=60$
Now from the figure, we can say that $BD=80-60=20$
So, the height of the first tree is 20 m.
Note: Be careful about the trigonometric ratios which you are using for the question. We have used the tangent of the given angle because we have to find the height and we have a common base in two right angle triangles that can be easily cancelled.
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