As in the last section we learned about the basics of trigonometry, in this section we will continue further with two topics i.e., **Complementary Angles and Heights and Distance**. In the first part we will first cover complementary angles and in the next heights & distances. So, let’s start with the first part.

__Complementary angles__**: – ** Do you know what complementary angles are? If yes, then well and good but if no, not an issue. In few seconds you will know what they are and there will be no difference between the one who knows and you {LOL}

Two angles are said to be the complement each other if their sum is 90^{o}.

If A and B are said to be complementary to each other if their sum is 90^{o}.

Simple, isn’t?

In trigonometry, trigonometry identities have special relations between them in terms of complementary. So what’s waiting for, let us begin!!

1. sinθ = cos ( 90^{o} – θ)

2. cosθ = sin ( 90^{o} – θ)

3. tanθ = cot ( 90^{o} – θ)

4. cotθ = tan ( 90^{o} – θ)

5. secθ = cosec ( 90^{o} – θ)

6. cosecθ = sec ( 90^{o} – θ)

We see here that sinθ and cosθ, tanθ and cotθ and secθ and cosecθ are complementary to each other.

Let’s solve an example to have a better understanding of the complementary angles. Here we go…

**Ques:-** Evaluate cos27^{o}/ sin63^{o} + sin70^{o}/ cos20^{o}.

**Solution:-** We know that sin and cos are complementary to each other and we also see that 27^{o} + 63^{o} = 90^{o} and 70^{o} + 20^{o} = 90^{o}

Hence, cos27^{o} = sin( 90^{o} – 27^{o}) = sin63^{o}

Likewise, sin70^{o} = cos (90^{o} – 20^{o}) = cos20^{o}

Therefore,

cos27^{o}/ sin63^{o} + sin70^{o}/ cos20^{o}

- sin63
^{o}/ sin63^{o}+ cos20^{o}/ cos20^{o}= 1 + 1 = 2

Hence, you see it was quite easy to calculate.

We can solve it in another way as well and it won’t make any difference as we have two complementary angles.

We could have written –

cos27^{o}/ sin63^{o} + sin70^{o}/ cos20^{o}

= cos27^{o}/ cos27^{o} + sin70^{o}/ sin70^{o} = 1+ 1 = 2

So, you can solve either ways it makes no difference.

As you now know about complementary angles, it would be easy for you to know what supplementary angles are.

__Supplementary angles__**:-** Two angles are said to be supplementary to each other if their sum is 180^{o}.

Two angles are said to be the complement each other if their sum is 180^{o}.

In ** SSC-CGL EXAM** questions are asked indirectly to make you confuse but if you have the right approach everything will be normal. So, let us look at some questions asked in the exam.

**Ques:- **The supplement of an angle is one – fourth of itself. Determine the angle and its supplement.** ( SSC- CGL)**

**Solution:- **As we have been asked about the supplementary angles. Let us assume that the first angle is x^{o} and the second will be 180^{o} – x^{o}.

According to the question,

Supplement of an angle = ¼ of itself

If we have x as an angle so its supplement will obviously be 180 – x.

Therefore Mathematically,

180^{o} – x^{o} = ¼ of x^{o}

- 4(180
^{o}– x^{o}) = x^{o} - 720
^{o}– 4x^{o}= x^{o} - 720
^{o}= 5x^{o} - x
^{o}= 144^{o}

So, our required angle is 144^{o} and its supplement will be 180^{o} – 144^{o} = 36^{o}.

As you can see there was no hard and core method or trick to solve such a question. That is why they are easily scoreable.

Let’s look at the second type of such questions and what actually is asked!

**Ques:- **If two supplementary angles are in the ratio 1:5. Find the difference between the angles.

**Solution:- **We’ll just go the way we did the previous question but here we have been given the ratio of the two angles. Let x and 5x be the two supplementary angles. If they are supplementary means that their sum is 180^{o}.

Therefore,

x^{o} + 5x^{o} = 180^{o}

- 6x
^{o}= 180^{o} - x
^{o}= 30^{o}

The first angle is, therefore, 30^{o} so the other angles will be 5 × 30 = 150^{o}

We have to find the difference between the two angles, so,

Required difference = 150^{o} – 30^{o} = 120^{o}.

As now you must be clear about the complementary and supplementary angles. Let us now move to the second part of this section i.e., heights and distances.

__HEIGHTS AND DISTANCE__

Heights and distances is one of the most important topics in this trigonometry section. It is important to learn the basics of this topic to have a clear picture as per the SSC-CGL EXAM perspective.

So guys first refresh your mind and then we begin with our learning.

Have you ever seen on the ground from the top of your terrace or any building? If not, then go run and see. Wait, wait, wait I’m not asking you to jump okay? (hahaha). I’m just asking you this for you to know that whenever you see from a top of building/ tower/ terrace/ pole, the angle you make with the ground to any object lying on it is known as ** Angle of Depression. **Likewise, if you are standing on the ground and you see an object above you is called the

Let me explain this with the help of a diagram.

Let AB is a building and you are at point C. on the ground who is viewing at C.

So, the angle which line makes with horizontal line BC is called the angle of elevation. So, angle ACB is the angle of elevation.

Similarly, when we talk about the angle of depression. We have,

So, here the angle from the horizontal downward to an object is the **angle of depression.**

**Note:-**

**1. **In case of the angle of depression – The observer’s line of sight would be below the horizontal.

**2. **In case of the angle of elevation – The observer’s line of sight would be above the horizontal.

**3. ** Numerically, the angle of elevation = angle of depression.

As now you must be clear with the concepts, let us solve some questions asked in previous years __SSC-CGL QUANT.__

**Ques:- **The angle of elevation of the top of a tower from the point P and Q at the distance of ‘a’ and ‘b’ respectively from the base of the tower and in the same straight line with it are complementary. Calculate the height of the tower. (__SSC- CGL, TIER II__)

**Solution:- **Let AB = h be the height of the tower.

According to question P and Q are on the same straight line are complementary. As we know, the sum of the complementary angle is 90. So, if Q makes θ angle then P makes 90 – θ.

In these types of question what we must compulsory do is make a rough sketch or diagram for our own clearance. Let us proceed to this-

*We have to find height*

*So, in triangle ABP,*

*tan(90 – θ) = h/ a*

*We know that tan (90 – θ) = cotθ*

*cotθ = h/ a**tanθ = a/ h ———–(1)*

*In triangle ABQ*

*tanθ = h/ b ———(2)*

*from (1) & (2)*

*a/ h = h/ b*

*h*^{2}= ab*h =*√ab

*So, the height of the tower is *√ab units.

So, this was a general form of such type of question, if you are asked in numerical terms all you need to do is substitute the value of a and b.

Let’s look at the question which is frequently asked in the exam.

**Ques:- **From two points on the ground lying on the straight line through the foot of a pillar, the two angles of elevation of the top of the pillar are complementary to each other. If the distance of the two points from the foot of the pillar is 9 meters and 16 meters and the two points lie on the same side of the pillar. Then find the height of the pillar.

**Explanation:- **If you observe this question and the last question we did you may see that there is not much difference to the solution. The only difference here is that instead of the constants ‘a’ and ‘b’ we have been given the numerical value of a and b.

Here, a = 9 and b = 16.

We need not calculate everything asked in the question, what will do is just substitute the value of a and b, to find the height.

So, height of the pillar = √ab = √9 × 16 = √144 = 12 metres.

Thus, you can see that once you know the concept of such type of questions, you only need to substitute the value as the question type is MCQ.

**Ques:- ** The top the two poles 24m and 36m are connected by the wire. If the wire makes an angle of 60^{o} with the horizontal, then find the length of the wire.__ (SSC-CGL,2017)__

**Solution/ explanation:-**

**Note:- ** this is a type II problem, by type-II I mean about the difficulty level.

So, let’s begin with our calculation-

As I always say we should first draw the diagram. Don’t wait for me to draw (hihi), draw it.

*Here, AB and CD are the two poles of length 24m and 36m respectively.*

*Let us take triangle AEC,*

*Here, sin60 ^{o} = AE/ AC*

- √3/2 = 12/ AC ——(1)

*We see in the diagram, AB = AE + EB*

*So, AE = AB – EB = 36 – 24 = 12m*

*AC = 12 × 2 / *√3 = 24/ √3 = 8 √3m

Therefore, the length of the wire = 8 √3m.