__Sine and Cosine Rules__See the image above and mug the formulas thoroughly. Sin and cosine rules are important in trigonometry and can help you in solving some complex questions.

The figure will look something like this -

In triangle ACD

sin∠CAD/3x = sin45/AD ... (1)

In triangle ABD

sin∠BAD/x = sin60/AD ... (2)

Divide equation (2) by (1)

3* sin∠BAD/sin∠CAD = sin60/sin45

sin∠BAD/sin∠CAD = √6/6 = 1/√6

**Answer: (C)**

**Similarly you can use cosine law to find any angle, if all the sides are given or to find a side, if the other two sides and an angle is given.**

**Q. 2) In the below figure, ABC is right angled at B and AD = CD. If ∠ACB=30, find ∠ABD**

**(A) 30 (B) 60 (C) 45 (D) 75**

**∠BAD = 180 - (90 + 30) = 60**

In triangle ABD

sin∠ABD/AD = sin60/BD [Since ∠BAD = 60]

In triangle BCD

sin∠CBD/CD = sin30/BD

Divide equation (2) by (1)

sin∠CBD/sin∠ABD = sin30/sin60 [AD = CD and hence they will cancel out]

Now, sin∠CBD = sin(90 - ∠ABD) = cos∠ABD [Click here if you don't know how to convert sin into cos]

Hence, cos∠ABD/sin∠ABD = sin30/sin60

cot∠ABD = 1/√3

Hence ∠ABD = 60

**Answer: (B)**

**Alternative Method**
The figure given in this question is very important and at times it is embedded in some other figures. There is one short-cut to calculate the angle.

Imagine the triangle in circumscribed in a circle

Now, D will be the centre of the triangle with diameter AC. We can say with surety that AC is the diameter of the circle because ∠ABC = 90, and we know angle in a

**semicircle**is right angle. Moreover AD = CD, hence D is the midpoint of the diameter or the centre of the circle
Now you can see that AD, CD and BD are radii of the circle. Hence AD = CD = BD

∠BAD = 180 - (90 + 30) = 60

∠ABD = ∠BAD = 60 [Since AD = BD]

**Answer: (B)**

**Now let us see a CGL question, in which the above figure was embedded.**

**Q. 3) G is the centroid of Triangle ABC, and AG=BC. Find angle BGC.**

**(A) 60 (B) 90 (C) 120 (D) 75**

Then BG = x (centroid divided the median in 2:1 ratio)

BC = AG = 2x

Let AG when extended cuts BC at D

Then D is the midpoint of BC (as AD is the median)

BD = DC = x [Since BC = 2x]

Now DG = BD = DC = x

That means D is the centre of a circle with diameter BC and one of the radius as DG.

Hence BGC = 90 (angle in a semi-circle)

**Answer: (B)**

sec

^{2}x + tan^{2}x = 5/3
We know, sec

^{2}x – tan^{2}x = 1
Adding the above two equations

2sec

^{2}x = 8/3 or sec^{2}x = 4/3
secx = 2/√3

That means x = 30

cos2x = cos60 = 1/2**Answer: (C)**

__Method 2:__sec

^{2}x = 4/3

That means, cos^2x = 3/4 and sin^2x = 1 - 3/4 = 1/4

cos2x = cos^2x - sin^2x = 3/4 - 1/4 = 1/2

Multiply and Divide RHS by 2

(cosx - sinx)/(cosx + sinx) = (√3/2 - 1/2)/(√3/2 + 1/2)

Match RHS with LHS and you can easily see x = 30

**Answer: (A)**

__Method 2__
Cross multiply

(cosx - sinx)(√3 + 1) = (cosx + sinx)(√3 - 1)

Solve it and you will get, tanx = 1/√3

Hence x = 30

2y*cos45 - x*sin45 = 0

2y = x ... (1)

2x*sec45 - y*cosec45 = 3

2x - y = 3/√2

4y - y = 3/√2 [Put x = 2y]

y = 1/√2

Hence, x = 2y = √2

x^2 + 4y^2 = 2 + 4*(1/2) = 2 + 2 = 4

**Answer: (C)**

**Some important values to mug:**

- sin15 = (√3 - 1) /2√2
- cos15 = (√3 + 1) /2√2
- tan15 = 2 - √3
- cot15 = 2 + √3

These values are very important to solve some tricky trigonometric questions. Examples :

**Q. 7)**

Although you can solve this question with the direct formula, which I discussed in Geometry Tricks - 1. But let us assume, you forget that formula. In such cases, the values of sin15 and cos15 will come handy.

Let the perpendicular and base of the triangle be P and B, respectively.

sin15 = Perpendicular/Hypotenuse = P/100

P = sin15*100Similarly, B = cos15*100

Area = 1/2 * P * B = 1/2 * sin15 * 100 * cos15 * 100 = (√3 - 1) /2√2 * (√3 + 1) /2√2 * 100 * 100/2

Area = 100*100/8 = 1250

**Answer: (D)**

**Q. 8)**

**A) 0 B) 1 C) -1 D) 2**

**Put x = 15**

= cot15/(cot15 - cot45) + tan15/(tan15 - tan45)

**=**(2

**+**√3)/(1 + √3) + (2 - √3)/(1 - √3)

= 1

**Answer: (B)**

**Some candidates are asking me the book which I refer to write these articles. But, I can assure you that all these tricks are authored by me and you won't find them in any book or coaching...**

**If you have any doubt in this article, please drop a comment...**

**Keep Reading :)**

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Thank you sir....

ReplyDeletethank u sooooooo much sir...i am a big admirer of you waitng a lot for ur next post.plz post soon. thankxxxxxx a lotttt

ReplyDeleteThanks for the awesome tips and your valuable time

ReplyDeletePlease keep posting in the blog !!

Dear sir,

ReplyDeleteKindly post the previous part of the post at bottom, it will help a new commer.

Thanks a lot for your hardwork.

Sir! I am preparing for SSC and I didn't buy any book for Quant section. I am just going through your blog and videos on youtube. Should i buy a book for quant ?

ReplyDeleteYes...you should at least buy 1 book for Quant because it is the most important section of CGL

DeleteHi Prashant,

DeleteShould i go for an advance book for maths like Lucent higher mathematics and Paramount by Rakesh Yadav?

If yes, then shoukd i try and complete both or just one of them will be enough? Also which is the better of the two books.

Since I have read neither, I can't comment...

DeleteSir from which book u studied advance maths ??

DeleteThank you sir.

ReplyDeleteSir , I just wanted know that MEA was your first choice ??

ReplyDeleteyes...

DeleteSir, paper k pattern ki koi news aayi change hoga ya same rahega???

ReplyDeleteno news as of now...

DeleteGood morning sir,

ReplyDeleteWaiting for your next post.

Thank you

thanks for your efforts,and keep doing this,many aspirants will be benifited.

ReplyDeleteSir in Q no 6 & 8 why we have to put only x =45 and x=15 respectively and how to decide..?

ReplyDeletesin∠CBD = cos(90 - ∠ABD).....part of q.2....i think some printing error in it...cos should be sin

ReplyDeleteSome typing errors in method-2 q.4...pls check

Deletewhat errors?

DeleteIn q.2...it should be "sin(90-ABD)...shuldn't it??

ReplyDeleteThis comment has been removed by the author.

ReplyDeleteSir in the last ques How you came to know that only x=15 will be assumed. Sir any other trick..

ReplyDeleteQ.3)

ReplyDeleteIf AG=2x, BG=2x and GC=2x (Since 2:1 division of median) and

since AG=BC given,

then,

BC=BG=GC, which implies BGC is an equilateral triangle,

So angle BGC=60 degrees

But according to your method (which i think is also correct), BGC = 90 degrees,

I think that it is not possible to form such a triangle with the given condition (AG=BC)

Correct me if i am wrong. :)

The length of all the medians can't be taken equal...

Delete