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.
- 1)
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.
- 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 [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.
- 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)
- 4)
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
- 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
- 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:
- 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 :
- 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)
- 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…
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