### 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 45 ^{0}– 45^{0}– 90^{0} 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 = 45^{o }& angle of elevation from B = ∠TBO = 60^{o}

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 45^{0 }– 45^{0})

In the right triangle BTO, OB/OT = 1/ √3 (as opposite angles of these sides are 30^{0 }– 60^{0})

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 = 30^{0} & Angle AEF = 60^{0}

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 30^{0}-60^{0})

FD = 28.5√3

In the right triangle AFE, AF/FE = 28.5/FE =√3/1 (as opposite angles of these sides are 60^{0}-30^{0})

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?

Height of the building = DB = 20 meters & Angle DCB = 45^{0}

In the right triangle DBC, DB/BC = 1/1 (as opposite angles of these sides are 45^{0 }– 45^{0})

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 60^{0 }– 30^{0})

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 = 60^{0} & Angle DCB =45^{0}

In the right triangle DBC, DB/BC = 1/1 (as opposite angles of these sides are 45^{0 }– 45^{0}) or BC = DB

In the right triangle ABC, AB/BC =√3 /1 (as opposite angles of these sides are 60^{0 }– 30^{0})

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 = 60^{0} & Angle DBC = 30^{0}

In the right triangle ABC, AB/ BC = √3/1 or 50/ BC = or √3/1BC = 50/√3 (as opposite angles of these sides are 60^{0}– 30^{0})

In the right triangle DCB, DC/ BC = 1/√3(as opposite angles of these sides are 30^{0}– 60^{0})

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 = 30^{0}

AB = ED = height of poles

In the right triangle ABC, AB/BC =√3/1 (as opposite angles of these sides are 60^{0}– 30^{0})

or AB = BC√3 ……. (i)

In the right triangle EDC, ED/CD = 1/√ 3 = AB/ (80-BC) (as opposite angles of these sides are 30^{0}-60^{0})

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 = 45^{0} & Angle EAD = 60^{0 }

In the right triangle ABC, AB = BC or AB = BC = AD = 7m (as opposite angles of these sides are 45^{0}– 45^{0})

In the right triangle EAD, ED/AD =√ 3/1 or ED = AD√ 3 = 7 √ 3 (as opposite angles of these sides are 60^{0 }– 30^{0})

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.**

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