Sunday, 1 May 2016

[Q] Trigonometry Tricks - 3


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.

Q. 1)

The figure will look something like this -


In triangle ACD
sinCAD/3x = sin45/AD   ... (1)
In triangle ABD
sin∠BAD/x = sin60/AD    ...  (2)
Divide equation (2) by (1)
3* sin∠BAD/sinCAD = sin60/sin45
sin∠BAD/sinCAD = √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


Let AG = 2x
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)


Q. 4)

sec2x + tan2x = 5/3
We know, sec2x – tan2x = 1
Adding the above two equations
2sec2x = 8/3 or sec2x = 4/3
secx = 2/√3
That means x = 30
cos2x = cos60 = 1/2
Answer: (C)
Method 2:
sec2x = 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



Q. 5)

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

Q. 6)

Put θ = 45
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:
  1.  sin15 = (√3 - 1) /2√2
  2.  cos15 = (√3 + 1) /2√2
  3.  tan15 = 2 - √3
  4.  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*100
Similarly, 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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25 comments:

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

    ReplyDelete
  2. Thanks for the awesome tips and your valuable time

    Please keep posting in the blog !!

    ReplyDelete
  3. Dear sir,
    Kindly post the previous part of the post at bottom, it will help a new commer.
    Thanks a lot for your hardwork.

    ReplyDelete
  4. 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 ?

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

      Delete
    2. Hi Prashant,
      Should 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.

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

      Delete
    4. Sir from which book u studied advance maths ??

      Delete
  5. Sir , I just wanted know that MEA was your first choice ??

    ReplyDelete
  6. Sir, paper k pattern ki koi news aayi change hoga ya same rahega???

    ReplyDelete
  7. Good morning sir,
    Waiting for your next post.
    Thank you

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

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

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

    ReplyDelete
    Replies
    1. Some typing errors in method-2 q.4...pls check

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

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  12. This comment has been removed by the author.

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

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  14. Q.3)

    If 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. :)

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

      Delete