Height and Distance

Height and Distance is a chapter wherein we use the concepts of Trigonometry to find the height and distances of various objects.

That is how the height Mount Everest, and distance of planets and stars from the earth have been calculated. So, you can say that this is the application part of Trigonometry.

Consider the figure given below: Trigonometry

Here E is representing the eye of an observer and EX is the horizontal line drawn through E.

Line of sight

The line joining the eye to an object, is called the line of sight.

So, EO and EO’ are the line of sights, joining the eye of the observer to the objects placed above (O) and below (O’) respectively.

Angle of elevation

If an object is above the horizontal line passing through the eye, then the observer would have to look up to see the object. For example, the object O in the above figure.

In such a case, the angle that the line of sight (EO) makes with the horizontal line passing through the eye (EX), is called the angle of elevation (as observed from E).

So, in the above figure, ∠XEO is the angle of elevation.

Angle of depression

If an object is below the horizontal line passing through the eye, then the observer would have to look down to see the object. For example, the object O’ in the above figure.

In such a case, the angle that the line of sight (EO’) makes with the horizontal line passing through the eye (EX), is called the angle of depression (as observed from E).

So, in the above figure, ∠XEO’ is the angle of depression.

Important Cases related to Height & Distance

Now, let us have a look at some important cases/scenarios that we will often encounter in aptitude examinations.

If we are already familiar with them, then that will save us some calculation and time in the exam hall.

Case 1: 30°, 60°, 90°

In a right angled triangle ABC, having angles 30°, 60° and 90°, the ratio of the length of the sides corresponding to (opposite to) these angles will be 1 : $\sqrt{3}$ : 2. Trigonometry

Case 2: 45°, 45°, 90°

In a right angled triangle ABC, having angles 45°, 45° and 90°, the ratio of the length of the sides corresponding to (opposite to) these angles will be 1 : 1 : $\sqrt{2}$. Trigonometry

Case 3

When two right angled triangles ABC and ABD have the same base, and apart from the common 90°, the other two angles are:

  • 30°, 60° in triangle ABC
  • 45°, 45° in triangle ABD

Then the ratio of AB : BD : AD : BC : AC = 1 : 1 : $\sqrt{2}$ : $\sqrt{3}$ : 2. Trigonometry We can see that, the ratio of BC : BD : DC = $\sqrt{3}$ : 1 : $\sqrt{3} - 1$

Case 4

When two right angled triangles ABC and ABD have the same base, and apart from the common 90°, the other two angles are:

  • 45°, 45° in triangle ABC
  • 30°, 60° in triangle ABD

Then the ratio of AB : BD : AD : BC : AC = $\sqrt{3}$ : 1 : 2 : $\sqrt{3}$ : $\sqrt{6}$. Trigonometry We can see that, the ratio of BC : BD : DC = $\sqrt{3}$ : 1 : $\sqrt{3} - 1$

Case 5

When two right angled triangles ABC and ABD have the same base, and apart from the common 90°, the other two angles are:

  • 60°, 30° in triangle ABC
  • 30°, 60° in triangle ABD

Then the ratio of AB : BD : AD : BC : AC = $\sqrt{3}$ : 1 : 2 : 3 : $2\sqrt{3}$. Trigonometry We can see that, the ratio of BC : BD : DC = 3 : 1 : 2

Case 6

When two right angled triangles ABC and ABD have the same base, and apart from the common 90°, the other two angles are:

  • 75°, 15° in triangle ABC
  • 60°, 30° in triangle ABD

Then the ratio of AB : BD : AD : BC= 1 : $\sqrt{3}$ : 2 : $\sqrt{3} + 2$. Trigonometry We can see that, the ratio of BC : BD : DC = $\sqrt{3} + 2$ : $\sqrt{3}$ : 2

Case 7

Cconsider a person who is standing at the bottom of a mountain, and is about to go up. Trigonometry

  • The angle of elevation of the person’s eyes to the summit of the mountain is 45°, when he is at the bottom of that mountain.
  • After ascending ’d’ distance towards the mountain at a slope of 30°, the angle of elevation to summit becomes 60°.

Then, the height of mountain, h = $\frac{d}{2} (\sqrt{3} + 1)$

Case 8

Consider a straight/vertical pole AB, and two points C and D on the ground. Two scenarios are possible, as shown below: Trigonometry

  • Distance of point C from the bottom of the pole, BC = c. And, the angle of elevation of the top of the pole from point C is θ°.
  • Distance of point D from the bottom of the pole, BD = d. And, the angle of elevation of the top of the pole from point D is Φ°.

If θ and Φ are complementary to each other, i.e. θ + Φ = 90°, then:
The height of the pole, h = $\sqrt{cd}$

Case 9

Consider two straight/vertical poles AB and CD, both standing on the same ground. Trigonometry

  • Height of pole AB is $h_1$, and height of pole CD is $h_2$.
  • Distance between the two poles is x units.
  • Angle of elevation of the top of pole AB from bottom of pole CD is θ°.
  • Angle of elevation of the top of pole CD from bottom of pole AB is Φ°.

If θ and Φ are complementary to each other, i.e. θ + Φ = 90°, then:
The distance between the two poles, x = $\sqrt{h_1 × h_2}$

Case 10

Consider a spherical balloon of radius r, hanging above an observer. Trigonometry

  • The balloon subtends an angle of α at the eye of an observer.
  • The angle of elevation of the center of the balloon (from the eye of the observer) is β.

Then, height of the center of the balloon from the ground (i.e. CD), h = r sin β × cosec $\frac{α}{2}$

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Last updated on Jan 8, 2022

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