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__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…**

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

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