### 18 Mar

Any candidate involved in SSC CGL preparation must pay special attention to the trigonometry part that is there in quantitative aptitude section. Within trigonometry, height & distance is one topic that you can’t afford to neglect as it can easily fetch you 3 to 4 valuable marks. But, if you lack the understanding of this topic then cracking exam questions can be a challenge. Moreover, cracking the questions alone isn’t sufficient. You should be good enough to arrive at the answer in less than a minute.

Keeping this in mind, the institute for Top SSC Coaching in South Delhi has explained out here the concepts & tricks related to this topic.

Basic Concepts

(I) Angle of Elevation – It is always measured from the ground up and is inside the triangle.

(II) Angle of Depression – When an object is below the observer, the angle made between the horizontal line and the line of sight of the observer is known as the angle of depression.
1. The lengths of the sides of a 450– 450– 900 triangle are in the ratio of 1: 1: √2.

2. The lengths of the sides of a 30°- 60°- 90° triangle are in the ratio of 1: √3: 2
Example 1: Angles of elevation of the top of a tower 30 meters high, from 2 points on the level ground on its opposite sides are 45 degrees and 60 degrees. Find the distance between these 2 points.

Solution: 47.32 meters
Let OT be the tower

Therefore, height of the tower = OT = 30 meters

Let A and B be the two points on the level ground on opposite sides of the tower OT.

Then, angle of elevation from A = ∠TAO = 45o & angle of elevation from B = ∠TBO = 60o

Distance between the two points (A & B) = AB = AO + OB = x + y (say)

In the right triangle ATO, AO = OT = x = 30 meters (as opposite angles of these sides are 450 – 450)

In the right triangle BTO, OB/OT = 1/ √3 (as opposite angles of these sides are 300 – 600)

or OB = OT/√3 = 30/√3 = 30√3/3 = 10√3 = 10×1.732 = 17.32

Hence, the required distance = x + y = 30 + 17.32 = 47.32 meters

Example 2: A boy who is 1.5 meters tall stands at some distance from a 30 meters tall building. When he starts walking towards the building, the angle of elevation from his eyes to the top of this building goes up from 30° to 60°. What is the distance walked by him towards the building?
Solution: 19√3 meters

AB = 30 meters (height of the building)

DC = EG = BF = 1.5 meters (height of the boy)

Angle ADF = 300 & Angle AEF = 600

AF = AB – BF = 30 – 1.5 = 28.5 meters

In the right triangle AFD, AF/FD = 1/√3= 28.5/FD (as opposite angles of these sides are 300-600)

FD = 28.5√3

In the right triangle AFE, AF/FE = 28.5/FE =√3/1  (as opposite angles of these sides are 600-300)

FE = 28.5/√3

Required distance = ED = FD – FE= 28.5√3- 28.5/√3= 28.5 (3 – 1)/√3

Example 3: From a point on the land, the angles of elevation of the bottom and the top of a television tower set at the top of a 20 meter tall building are 45° and 60° respectively. What is the height of the tower?

Solution: 14.68 meters

Height of the building = DB = 20 meters & Angle DCB = 450

In the right triangle DBC, DB/BC = 1/1 (as opposite angles of these sides are 450 – 450)

Or DB = BC = 20 meters

In the right triangle ABC, AB/BC= AB/ 20 =√3/1  or AB = 20√3 (as opposite angles of these sides are 600 – 300)

Now, AD = AB – DB = 20√3-20  = 20×1.732 – 20 = 34.68 – 20 = 14.68 meters

Example 4: A statue of height 1.6 meters stands at the top of a pedestal. From a point on the land, the angle of elevation at the top of this statue is 60° and from this point itself, the angle of elevation at the top of the pedestal is 45°. What is the height of this pedestal?
Solution: 21.85 meters

Height of the statute = AD = 1.6 meters

Height of the pedestal = BD =?

Angle ACB = 600 & Angle DCB =450

In the right triangle DBC, DB/BC = 1/1 (as opposite angles of these sides are 450 – 450) or BC = DB

In the right triangle ABC, AB/BC =√3 /1 (as opposite angles of these sides are 600 – 300)

or BC = AB / √3 = (BD +16)/ = (BC +16)/√3

or BC √3 –  BC =  16

or BC(√3-1) = 16 or BC ×0.732 = 16

or BC =  16/0.732 = 21.85 meters

Example 5: The angle of elevation at the top of a building from the base of a tower is 30° & the angle of elevation at the top of the tower from the base of the building is 60°. If the height of the tower is 50 meters, then what is the height of the building?
Solution: 50/3 meters

Height of the tower = AB = 50 meters

Height of the building = DC =?

Angle ACB = 600 & Angle DBC = 300

In the right triangle ABC, AB/ BC = √3/1 or 50/ BC =  or √3/1BC = 50/√3  (as opposite angles of these sides are 600– 300)

In the right triangle DCB, DC/ BC = 1/√3(as opposite angles of these sides are 300– 600)

or DC = BC/√3 = ( 50/√3/√3m = 50/3 meters

Example 6: Two poles of the same height stand opposite to each other on either side of a road that is 80 meters wide. From a particular point between them on the road, the angles of elevation at the top of these poles are 60° and 30° respectively. What are the heights of the poles and the distances of this point from these poles
Solution: Height = 20√3 meters & Distances = 20 meters & 60 meters.

BD = width of road = 80 meters

Angle ACB = 60 & Angle ECD = 300

AB = ED = height of poles

In the right triangle ABC, AB/BC =√3/1  (as opposite angles of these sides are 600– 300)

or AB = BC√3   ……. (i)

In the right triangle EDC, ED/CD = 1/√ 3  = AB/ (80-BC) (as opposite angles of these sides are 300-600)

Or AB = (80 – BC)/√3  …… (ii)

Or BC  = (80 – BC)/√3  from …. (i)

Or 3 BC = 80 – BC

Or 4 BC = 80

Or BC = 20 meters & CD = 60 meters

So, AB = 20√3 meters

Example 7: From the top of a 7 meter tall building, the angle of elevation at the top of a transmission tower is 60° and the angle of depression of its base is 45°. Find the height of this tower.
Solution: 19.124 meters

AB = height of the building = 7 meters

Angle DAC = ACB = 450 & Angle EAD =  60

In the right triangle ABC, AB = BC or AB = BC = AD = 7m (as opposite angles of these sides are 450– 450)

In the right triangle EAD, ED/AD =√ 3/1 or ED = AD√ 3 =   7 √ 3 (as opposite angles of these sides are 600 – 300)

So, height of the tower = DC + ED = 7 + 7√3 = 7(1+√3) = 7(1+1.732) = 7×2.732 = 19.124 meters.

Finally, the concepts discussed above are equally important even if you are doing SSC CHSL Coaching in Delhi; trigonometry is there on the 10+2 / LDC level exam as well. But, you must make it a point to attempt a lot of practice questions as well. Doing so will improve your speed and also result in greater accuracy.

Summary

This informative article on Trigonometry is the result of efforts put in by Vidya Guru Institute experts. For clarification of any doubts & queries, you can write to vidyagurudelhi@gmail.com.