The angles of elevation of a tower from a point on the ground is ${30^ \circ }$. At a point on the horizontal line passing through the foot of the tower and 100 metres nearer to it. If the angle of elevation is found to be ${60^ \circ }$, then height of the tower is
A. $50\sqrt 3 $ metres
B. $\dfrac{{50}}{{\sqrt 3 }}$ metres
C. $100\sqrt 3 $ metres
D. $\dfrac{{100}}{{\sqrt 3 }}$ metres
Answer
605.4k+ views
Hint:
First draw a corresponding diagram in this question. Let the height of the tower be $h$ and the total distance from foot of tower to the initial point be $x$. Then, the distance from foot perpendicular to the second position is $100 - x$. Then, use the trigonometric ratios to form equations and hence solve the equations to find the value of $h$.
Complete step by step solution:
We will first draw a corresponding diagram of the given function.
Let $h$ be the height of the tower. Here, let the total distance from foot of tower to the initial point be $x$ and the total distance from foot of perpendicular to the second position be $100 - x$.
The angle of elevation of a tower from a point on the ground is ${30^ \circ }$ and after 100 metres, the angle of elevation is found to be ${60^ \circ }$.
Now, consider the triangle ABC,
$\tan {60^ \circ } = \dfrac{h}{{x - 100}}$
We will substitute the value of $\tan {60^ \circ }$ as $\sqrt 3 $
$\sqrt 3 = \dfrac{h}{{x - 100}}$ eqn. (1)
Consider the triangle ABD
$\tan {30^ \circ } = \dfrac{h}{x}$
On substituting the value of $\tan {30^ \circ }$ as $\dfrac{1}{{\sqrt 3 }}$
$\dfrac{1}{{\sqrt 3 }} = \dfrac{h}{x}$
The value of $x$ from the above equation, is $x = h\sqrt 3 $
Substitute the value of $x$ in equation (1)
$
\sqrt 3 = \dfrac{h}{{h\sqrt 3 - 100}} \\
\Rightarrow 3h - 100\sqrt 3 = h \\
\Rightarrow 2h = 100\sqrt 3 \\
$
Divide the equation by 2
$h = 50\sqrt 3 $
Hence, the height of the tower is $50\sqrt 3 $ metres.
Thus, option A is correct.
Note:
The angle of elevation is the angle between the horizontal line and to the height of the object. It is also known as upward angle. And the trigonometry ratio is the ratio of sine angle to the cosine of that angle.
First draw a corresponding diagram in this question. Let the height of the tower be $h$ and the total distance from foot of tower to the initial point be $x$. Then, the distance from foot perpendicular to the second position is $100 - x$. Then, use the trigonometric ratios to form equations and hence solve the equations to find the value of $h$.
Complete step by step solution:
We will first draw a corresponding diagram of the given function.
Let $h$ be the height of the tower. Here, let the total distance from foot of tower to the initial point be $x$ and the total distance from foot of perpendicular to the second position be $100 - x$.
The angle of elevation of a tower from a point on the ground is ${30^ \circ }$ and after 100 metres, the angle of elevation is found to be ${60^ \circ }$.
Now, consider the triangle ABC,
$\tan {60^ \circ } = \dfrac{h}{{x - 100}}$
We will substitute the value of $\tan {60^ \circ }$ as $\sqrt 3 $
$\sqrt 3 = \dfrac{h}{{x - 100}}$ eqn. (1)
Consider the triangle ABD
$\tan {30^ \circ } = \dfrac{h}{x}$
On substituting the value of $\tan {30^ \circ }$ as $\dfrac{1}{{\sqrt 3 }}$
$\dfrac{1}{{\sqrt 3 }} = \dfrac{h}{x}$
The value of $x$ from the above equation, is $x = h\sqrt 3 $
Substitute the value of $x$ in equation (1)
$
\sqrt 3 = \dfrac{h}{{h\sqrt 3 - 100}} \\
\Rightarrow 3h - 100\sqrt 3 = h \\
\Rightarrow 2h = 100\sqrt 3 \\
$
Divide the equation by 2
$h = 50\sqrt 3 $
Hence, the height of the tower is $50\sqrt 3 $ metres.
Thus, option A is correct.
Note:
The angle of elevation is the angle between the horizontal line and to the height of the object. It is also known as upward angle. And the trigonometry ratio is the ratio of sine angle to the cosine of that angle.
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