Exams Prep Master
The topic Heights and Distances is one of the important applications of trigonometry, which can be applied extensively in our real life. By using these trigonometric applications, astronomers, navigators, architects, and surveyors can solve problems related to heights and distances.
Trigonometry
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Trigonometry has found its importance in multiple fields. It’s found its applications within the fields starting from Engineering to Architecture to Astronomy also. Trigonometry is often utilized in these fields to live distances and angles by assuming lines that connect the points.
Height and Distance
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One among the most important applications of trigonometry is to seek out the space between two or quite two places or to seek out the peak of the thing or the angle subtended by any object at a given point without actually measuring the space or heights or angles. Trigonometry is beneficial to astronomers, navigators, architects and surveyors etc. in solving problems associated with heights and distances.
Height and Distance
Using trigonometry, if we are given any of the 2 quantities which will be a side or an angle, we will calculate all the remainder of the quantities. By the law of alternate angles, the angle of elevation and angle of depression are consequently equal in magnitude (α = β). Tan α is adequate to the ratio of the peak and distance.
You can be given any two of the subsequent information:
- The distance of the thing from the observer
- The height of the thing
- Angle at which the observer views the topmost point of the thing (angle of elevation)
- The angle at which the observer views the thing when the observer is on top of a tower/building (angle of depression)
Angle of elevation
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Angle of elevation
Let C be the position of the thing above the horizontal line AB and B be the attention of the observer, then angle ACB is named angle of elevation.
Angle of Depression
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The angle of depression is the angle between the horizontal line and the observation of the object from the horizontal line. It is basically used to get the distance of the two objects where the angles and an object’s distance from the ground are known to us.
It’s an angle that is formed with the horizontal line if the line of sight is downward from the horizontal line.
Here, in the below fig, the θ is the angle of inclination
∠ABO is the Angle of elevation and
∠O is the Angle of depression
Angle of Depression
Theodolite
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Angles of elevation or angles of depression of an object is measured by an instrument known as Theodolite. Theodolite is predicated on the principles of trigonometry, which is employed for measuring angles with a rotating telescope. In 1856, Sir George Everest first used the enormous theodolite, which is now on display within the Museum of the survey of India in Dehradun.
Theodolite
Sample Questions
Ques: What does height mean? [1 mark]
Ans. Basically, height means altitude or elevation that refers to the space above A level. Furthermore, height denotes extent upward (as from foot to head) also as any measurable distance above a given level. For instance, the peak of a tree, human, mountain, tower, etc.
Ques: What's the difference between height and height? [1 mark]
Ans. The main difference between height and high is that height refers to the space between the very best and lowest end of an object. On the opposite hand, high refers to a mountain located within the US of America.
Ques: How does one find the peak of objects in math? [1 mark]
Ans. In mathematics, you'll calculate the peak of an object using the space and angles. Here distance is that the horizontal distance between the thing and therefore the angle is that the angle above the horizontal of the highest of the thing, which provides the peak of the thing.
Ques: What's the utmost height? [1 mark]
Ans. It refers to the very best vertical position along the trail of its trajectory. Moreover, the utmost height of the projectile depends upon the speed and angle of the launch of the thing and therefore the acceleration thanks to gravity
Previous Years’ Questions
Ques: A ladder 15 m long, just reaches the top of a vertical wall. If the ladder makes an angle of 60 degree with the wall, then calculate the height of the wall. (2013 OD)
Ans.
Let AB be the wall and AC be the ladder,
Ques: In the given figure, a tower AB is 20 m high and BC is its shadow on the ground which is \(2\sqrt{3}\) m long. Find the sun’s altitude. (2015 OD)
Ans. AB = 20m, BC = \(2\sqrt{3}\) m and θ = ?
In Δ ABC,
\(\Rightarrow tan\Theta = \frac{20}{20\sqrt{3}}\)
\(\Rightarrow \Theta=tan^{-1}(\frac{1}{\sqrt{3}})\)
\(\Rightarrow \) \(\Theta = 30^{\circ}\)
Ques: A ladder leaning against a wall, makes an angle of 60° with the horizontal. If the foot of the ladder is 2.5 m away from the wall, what is the length of the ladder? (2016 OD)
Ans. Let AC be the ladder.
\(cos 60^{\circ}=\frac{BC}{AC}\)
\(\frac{1}{2}=\frac{2.5}{AC}\)
AC = 5m
Therefore, length of the ladder, AC = 5m.
Ques: If a tower 30 m high, casts a shadow 10√3 m long on the ground. Find the angle of elevation of the sun. (2017OD)
Ans. Let the required angle be θ
\(tan\Theta=(\frac{30}{10\sqrt{3}})\)
\(tan\Theta=\sqrt{3}\)
⇒ tan θ = tan 60°
Therefore, θ = 60°
Ques: The top of the two towers of height x and y, standing on level ground, suntend angle of 30° and 60° respectively at the centre of the line, joining their feet. Find x:y. (2015D)
Ans. 
When base is same for both the towers and their heights are given as x and y respectively,
Suppose E is the centre of the line joining the feet of the two towers
Now, in Δ ABE, \(\frac{AB}{BE}=tan30^{\circ}\)
\(BE=\sqrt{3x}\)...(1)
Now In Δ CDE, \(\frac{CD}{DE}=tan60^{\circ}\)
\(DE=\frac{y}{\sqrt{3}}\)...(2)
Now BE=DE ....(3) (E is mid-point of BD)
\(\frac{x}{y}=\frac{1}{3}\),
Hence, the ratio of x and y is 1 : 3.
Ques: The angles of depression of two ships from the top of a lighthouse and on the same side of it are found to be 45° and 30°. If the ships are 200 m apart, what is the height of the lighthouse? (2015D)
Ans. 
Therefore, the height of the lighthouse is 273.2 m.
Ques: The angle of the elevation of an aeroplane from a point on the ground is 60°. After a flight of 30 seconds, the angle of elevation becomes 30°. If the aeroplane is flying at a constant height of 3000\(\sqrt{3}\) m, what is the speed of the aeroplane? (2014OD)
Ans.
\(tan60^{\circ}=\frac{3000\sqrt{3}}{x}\)
\(x=\frac{3000\sqrt{3}}{\sqrt{3}}=3000m\)
\(tan30^{\circ}=\frac{3000\sqrt{3}}{\frac{1}{\sqrt{3}}}=9000m\)
distance covered = 9000- 3000
= 6000 m.
speed = \(\frac{6000}{30}=200m/s\)
Ques: The angles of depression of the top and bottom of a 50 m high building from the top of a tower are 45° and 60° respectively. What is the height of the tower and the horizontal distance between the tower and the building? [Use \(\sqrt{3}\) = 1. 73] (2016D)
Ans.
Let AE be the building and CD be the tower and let height of the tower be h m.
And the horizontal distance between tower and the building = x m …. (Given CD = AE = 50m )
\(h=75 + 25\sqrt{3}\)
The horizontal distance between the tower and the building, x = 118. 25 m.
Ques: From the top of the vertical tower, the angles of depression of two cars in the same straight line with the base of the tower, at an instant are found to be 45° and 60°. If the cars are 100 m apart and are on the same side of the tower, what is the height of the tower? [Use \(\sqrt{3}\) = 1. 73] (2011 OD)
Ans. 
Let AB=x,
In △ABC,
\(tan 60^{\circ}=\frac{AB}{BC}\)
\(Bc=\frac{x}{\sqrt{3}}\)
In △ABD,
\(tan 45^{\circ}=\frac{AB}{BC+100}\)
BC = x – 100
x = \(x\sqrt{3}\) − \(100\sqrt{3}\)
x (\(\sqrt{3}\)−1) = 100\(\sqrt{3}\)
\(x=\frac{(100)\sqrt{3}\times (\sqrt{3}+1)}{2}\)
= 236.60 m
Therefore, the height of the tower, h = 236.60 m.
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