
A vertical tower is 20 m high. A man standing at some distance from the tower knows that the cosine of angle of elevation of the top of the tower is 0.53. How far is he standing from the foot of the tower?
Answer
509.7k+ views
Hint: To solve this question we will use the trigonometric identity and the given value to determine cos. The formula to calculate cos where angle is given as \[\theta \] is \[\cos \theta =\dfrac{\text{base}}{\text{hypotenuse}}\].
Complete step-by-step answer:
Given that a vertical tower is 20 m high. A man standing at some distance from the tower knows that the cosine of angle of elevation of the top of the tower is 0.53.
Which implies that if the angle of elevation is \[\theta \], then we have cos θ = 0.53.
Now we have to determine the distance between the base of the tower and the place where the man is standing.
Let the distance of the man from the foot of the tower is x.
Let AB be the tower and C be the point where the man is standing. Then,
AB = 20m, and BC = x and AC = y
Now because the obtained triangle is a right angles triangle therefore, we apply Pythagoras theorem stated as In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides.
Then applying Pythagoras theorem in triangle ABC we get,
\[\begin{align}
& AC=\text{ }\sqrt{{{\text{x}}^{2}}+{{\left( \mathbf{20} \right)}^{\mathbf{2}}}}\text{ } \\
& \Rightarrow AC=y=\sqrt{{{\text{x}}^{2}}+400} \\
\end{align}\]
Now,
\[\cos \theta =\dfrac{\text{base}}{\text{hypotenuse}}\].
Applying this on the triangle ABC we get,
\[\begin{align}
& \Rightarrow cos\text{ }\theta =\dfrac{x}{\sqrt{{{\text{x}}^{2}}+400}} \\
& \Rightarrow cos\text{ }\theta =0.53\text{ }=\dfrac{x}{\sqrt{{{\text{x}}^{2}}+400}} \\
\end{align}\]
Squaring both sides, we get,
\[{{\left( 0.53 \right)}^{2}}=\dfrac{{{x}^{2}}}{x^2\text{ }+\text{ }400}\]
\[\Rightarrow 0.2809\text{ }=\text{ }\dfrac{{{x}^{2}}}{x^2\text{ }+\text{ }400}\]
\[\Rightarrow {{x}^{2}}\text{ }=\text{ }0.2809{{x}^{2}}\text{ }+\text{ }112.36\]
\[\Rightarrow \text{ }{{x}^{2}}\text{ }\text{ }- 0.2809{{x}^{2}}\text{ }=\text{ }112.36\]
\[\Rightarrow 0.7191{{x}^{2}}=112.36\]
\[\begin{align}
& \Rightarrow {{x}^{2}}=\dfrac{112.36}{0.7191} \\
& \Rightarrow x=12.5\text{ }m \\
\end{align}\]
Therefore, we obtained the value of x as 12.5m.
Hence, the distance between the base of the tower and the place where the man is standing is given by 12.5 m.
Note: The possibility of error in this question can be at the where we have to determine the length of the hypotenuse of the right-angled triangle. The question cannot be solved using Pythagoras theorem so don’t opt for shortcuts, try to go with full steps.
Complete step-by-step answer:
Given that a vertical tower is 20 m high. A man standing at some distance from the tower knows that the cosine of angle of elevation of the top of the tower is 0.53.
Which implies that if the angle of elevation is \[\theta \], then we have cos θ = 0.53.
Now we have to determine the distance between the base of the tower and the place where the man is standing.
Let the distance of the man from the foot of the tower is x.
Let AB be the tower and C be the point where the man is standing. Then,
AB = 20m, and BC = x and AC = y
Now because the obtained triangle is a right angles triangle therefore, we apply Pythagoras theorem stated as In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides.
Then applying Pythagoras theorem in triangle ABC we get,
\[\begin{align}
& AC=\text{ }\sqrt{{{\text{x}}^{2}}+{{\left( \mathbf{20} \right)}^{\mathbf{2}}}}\text{ } \\
& \Rightarrow AC=y=\sqrt{{{\text{x}}^{2}}+400} \\
\end{align}\]
Now,
\[\cos \theta =\dfrac{\text{base}}{\text{hypotenuse}}\].
Applying this on the triangle ABC we get,
\[\begin{align}
& \Rightarrow cos\text{ }\theta =\dfrac{x}{\sqrt{{{\text{x}}^{2}}+400}} \\
& \Rightarrow cos\text{ }\theta =0.53\text{ }=\dfrac{x}{\sqrt{{{\text{x}}^{2}}+400}} \\
\end{align}\]
Squaring both sides, we get,
\[{{\left( 0.53 \right)}^{2}}=\dfrac{{{x}^{2}}}{x^2\text{ }+\text{ }400}\]
\[\Rightarrow 0.2809\text{ }=\text{ }\dfrac{{{x}^{2}}}{x^2\text{ }+\text{ }400}\]
\[\Rightarrow {{x}^{2}}\text{ }=\text{ }0.2809{{x}^{2}}\text{ }+\text{ }112.36\]
\[\Rightarrow \text{ }{{x}^{2}}\text{ }\text{ }- 0.2809{{x}^{2}}\text{ }=\text{ }112.36\]
\[\Rightarrow 0.7191{{x}^{2}}=112.36\]
\[\begin{align}
& \Rightarrow {{x}^{2}}=\dfrac{112.36}{0.7191} \\
& \Rightarrow x=12.5\text{ }m \\
\end{align}\]
Therefore, we obtained the value of x as 12.5m.
Hence, the distance between the base of the tower and the place where the man is standing is given by 12.5 m.
Note: The possibility of error in this question can be at the where we have to determine the length of the hypotenuse of the right-angled triangle. The question cannot be solved using Pythagoras theorem so don’t opt for shortcuts, try to go with full steps.
Recently Updated Pages
Two men on either side of the cliff 90m height observe class 10 maths CBSE

What happens to glucose which enters nephron along class 10 biology CBSE

Cutting of the Chinese melon means A The business and class 10 social science CBSE

Write a dialogue with at least ten utterances between class 10 english CBSE

Show an aquatic food chain using the following organisms class 10 biology CBSE

A circle is inscribed in an equilateral triangle and class 10 maths CBSE

Trending doubts
Why is there a time difference of about 5 hours between class 10 social science CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

What is the median of the first 10 natural numbers class 10 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Which of the following does not have a fundamental class 10 physics CBSE

State and prove converse of BPT Basic Proportionality class 10 maths CBSE

