Quantitative Aptitude PYQs

Concept:

Cosine formula is given by:

cos C = \(\frac{a^2+b^2-c^2}{2ab}\) for the triangle shown in the figure.

Calculation:

The figure drawn from the information given in the question is :

By cosine formula ,

cos C = \(\frac{a^2+b^2-c^2}{2ab}\) , where a = 10cm, b = 10cm and c = 16cm.

Thus, cos C = \(\frac{10^2+10^2-16^2}{2(10)(10)}\)

⇒cos C = -\(\frac{7}{25}\)

⇒sin C =\(\sqrt{1- cos^2C}\) = \(\sqrt{1-(-\frac{7}{25})^2}\)

⇒sin C = \(\sqrt{1-\frac{49}{625}}\)

⇒sin C = \(\sqrt{\frac{576}{625}}=\frac{24}{25}\)

Thus, \(\frac{AB}{sin C}=\frac{16}{\frac{24}{25}}=\frac{16\times25}{24}=\frac{50}{3}\)

⇒\(\frac{AB}{sin C}=\frac{50}{3}\)

Concept:

For any triangle ABC, (as shown in the figure)

cosine formula is given by:

cos C = \(\frac{a^2+b^2-c^2}{2ab}\) 

Calculation:

We need to find cos A + cos B+ cos C.....(equation 1)

It can be found by finding cos A, cos B and cos C respectively.

Thus,

By cosine formula ,

cos C = \(\frac{a^2+b^2-c^2}{2ab}\) , where a = 10cm, b = 10cm and c = 16cm.

Thus, cos C = \(\frac{10^2+10^2-16^2}{2(10)(10)}\)

⇒cos C = -\(\frac{7}{25}\)

Now, finding cos B = \(\frac{a^2+c^2-b^2}{2ac}\),where a = 10cm, b = 10cm and c = 16cm.

Thus, cos B = \(\frac{10^2+16^2-10^2}{2(10)(16)}\)

⇒cos B = \(\frac{4}{5}\)

Now, finding cos A = \(\frac{b^2+c^2-a^2}{2bc}\),where a = 10cm, b = 10cm and c = 16cm.

Thus, cos B = \(\frac{10^2+16^2-10^2}{2(10)(16)}\)

⇒cos A  = \(\frac{4}{5}\)

Thus, putting cos A, cos B and cos C in equation 1, we get

cos A + cos B + cos C = \(\frac{4}{5}+\frac{4}{5}-\frac{7}{25}\)

⇒cos A + cos B + cos C = \(\frac{33}{25}\)

Concept:

tan(x + y) = \(\frac {tan x +tany}{1-(tanx)(tany)}\)

tan(x - y) = \(\frac {tan x -tany}{1+(tanx)(tany)}\)

Application:

According to the question, the tower leaning towards north and has two points P and Q subtending an angle of 15° and 45° respectively(as can be seen in the figure)

Let the distance between OA be a 

and We have to find height of the top of the tower from the ground level i.e. OM.

Let OM be m.

In the Triangle MOP,

tan(15°) = \(\frac{OM}{OP}\)

⇒tan(45° - 30°) = \(\frac{tan45°-tan30°}{1+tan45° tan30° }\)=\(\frac{OM}{OP}\)

⇒\(\frac{OM}{OP}\) =  \(\frac{1-\sqrt3}{1+\sqrt3}\)   

⇒\(\frac{OM}{OP}=\frac{1-\sqrt3}{1+\sqrt3}[\frac{1-\sqrt3}{1-\sqrt3}]\)(Rationalizing)

⇒\(\frac{OM}{OP}=\frac{(1-\sqrt3)^2}{1^2-\sqrt3^2}\)

⇒\(\frac{m}{a+x}=\frac{1-2\sqrt3+3}{1-3}\)

⇒\(\frac{m}{a+x}=\sqrt3-2\)

⇒m = (a + x)(\(\sqrt3\) - 2)..........(equation 1)

Now, considering triangle MOQ,

tan(75°) = \(\frac{OM}{OQ}\)

⇒tan(45° + 30°) = \(\frac{tan45°+tan30°}{1-tan45° tan30° }\) = \(\frac{OM}{OQ}\)

⇒\(\frac{OM}{OQ}=\frac{1+\sqrt3}{1-\sqrt3}\)

⇒\(\frac{OM}{OQ}=\frac{1+\sqrt3}{1-\sqrt3}[\frac{1+\sqrt3}{1+\sqrt3}]\) (Rationalizing)

⇒\(\frac{OM}{OP}=\frac{(1+\sqrt3)^2}{1^2-\sqrt3^2}\)

⇒\(\frac{m}{a+y}=\frac{1+2\sqrt3+3}{1-3}\)

⇒\(\frac{m}{a+y}=-(\sqrt3+2)\)

⇒ m = - (a + y)(\(\sqrt3+2\))

Now, putting the value of m from equation1, we get

 (a + x)(\(\sqrt3\) - 2)= - (a + y)(\(\sqrt3+2\))

⇒\(\sqrt3\)a - 2a +\(\sqrt3\)x - 2x = - (\(\sqrt3\) a + 2a + \(\sqrt3\) y + 2y) 

⇒(\(\sqrt3\)a - 2a + \(\sqrt3\) a + 2a) = 2x - 2y - \(\sqrt3\)x - \(\sqrt3\)y

⇒ 2(\(\sqrt3\) )a = 2(x - y) - (\(\sqrt3\) )(x + y)

⇒a =   \(\frac{2(x-y)-( \sqrt3)(x+y)}{2\sqrt3}\)

Putting this value of a in equation 1,

m = ( \(\frac{2(x-y)-( \sqrt3)(x+y)}{2\sqrt3}\) + x)(\(\sqrt3\) - 2)

⇒m = \(\frac{2(x-y)-( \sqrt3)(x+y)+2\sqrt3x}{2\sqrt3}\)(\(\sqrt3\) - 2)

⇒m =  \(\frac{x(2+\sqrt3)-y(2+\sqrt3)}{2\sqrt3}\)(\(\sqrt3\) - 2)

On Simplifying, we get m = \(\frac{x-y}{2\sqrt3}\)

Solution:

Height of the tower AB = h

Distance between first station to foot of tower PA = x + a

Distance between second station to foot of tower QA = y + a

Distance between C and A = a

Give, 15^o, 75^o are the angle of elevation two stations to top of the tower

that is ∠APB = 15^o, ∠AQB = 75^o, ∠ACB = θ 

In \(\Delta PAB \) 

\(cot15^o =\frac{x+a}{h}\)

x + a = hcot15^o

Multiply y on both sides

xy + ay = hycot15^o ....(1)

In \(\Delta QAB \)

\(cot75^o =\frac{y+a}{h}\)

y + a = hcot75^o 

Multiply x on both sides

xy + ax = hxcot75^o .....(2)

In \(\Delta CAB \)

\(cot\theta =\frac{a}{h}\) .....(3)

Substract (2) - (1)

a(x - y) = h(xcot75^o - ycot15^o)

\(\frac{a}{h}=\frac{xcot75^{o}-ycot15^{o}}{x-y}\)

\(cot\theta =\frac{xcot75^{o}-ycot15^{o}}{x-y}\)

\(cot\theta =\frac{x(2-\sqrt{3})-y(2+\sqrt{3})}{x-y}\)

\(cot\theta =\frac{2x-\sqrt{3}x-2y-\sqrt{3}y}{x-y}\)

\(cot\theta =\frac{2(x-y)-\sqrt{3}(x+y)}{x-y}\)

\(cot\theta =2-\frac{\sqrt{3}(x+y)}{x-y}\)

∴ The correct option is (4)

Concept:

tan(x + y) = \(\frac {tan x +tany}{1-(tanx)(tany)}\)

tan(x - y) = \(\frac {tan x -tany}{1+(tanx)(tany)}\)

Calculation:

According to the question, the tower leaning towards north and has two points P and Q subtending an angle of 15° and 45° respectively(as can be seen in the figure)

Let the distance between OA be a 

and We have to find length of the tower i.e  AM

But before that, lets find the height of the tower above the ground level i.e. OM

Let OM be m.

In the Triangle MOP,

tan(15°) = \(\frac{OM}{OP}\)

⇒tan(45° - 30°) = \(\frac{tan45°-tan30°}{1+tan45° tan30° }\)=\(\frac{OM}{OP}\)

⇒\(\frac{OM}{OP}\) =  \(\frac{1-√3}{1+√3}\)   

⇒\(\frac{OM}{OP}=\frac{1-√3}{1+√3}[\frac{1-√3}{1-√3}]\)(Rationalizing)

⇒\(\frac{OM}{OP}=\frac{(1-√3)^2}{1^2-√3^2}\)

⇒\(\frac{m}{a+x}=\frac{1-2√3+3}{1-3}\)

⇒\(\frac{m}{a+x}=√3-2\)

⇒m = (a + x)(√3 - 2)..........(equation 1)

Now, considering triangle MOQ,

tan(75°) = \(\frac{OM}{OQ}\)

⇒tan(45° + 30°) = \(\frac{tan45°+tan30°}{1-tan45° tan30° }\) = \(\frac{OM}{OQ}\)

⇒\(\frac{OM}{OQ}=\frac{1+√3}{1-√3}\)

⇒\(\frac{OM}{OQ}=\frac{1+√3}{1-√3}[\frac{1+√3}{1+√3}]\)  (Rationalizing)

⇒\(\frac{OM}{OP}=\frac{(1+√3)^2}{1^2-√3^2}\)

⇒\(\frac{m}{a+y}=\frac{1+2√3+3}{1-3}\)

⇒\(\frac{m}{a+y}=-(√3+2)\)

⇒m = - (a + y)(\())

Now, putting the value of m from equation1, we get

 (a + x)(\(√3\) - 2)= - (a + y)(\(√3+2\))

⇒\(√3\)a - 2a +\(√3\)x - 2x = - (\(√3\) a + 2a + \(√3\) y + 2y) 

⇒(\(√3\)a - 2a + \(√3\) a + 2a) = 2x - 2y - \(√3\)x - \(√3\)y

⇒2(\(√3\) )a = 2(x - y) - (\(√3\) )(x + y)

⇒a =   

Putting this value of a in equation 1,

m = ( \(\frac{2(x-y)-( √3)(x+y)}{2√3}\) + x)(\(√3\) - 2)

⇒m = \(\frac{2(x-y)-( √3)(x+y)+2√3x}{2√3}\)(\(√3\) - 2)

⇒m =  \(\frac{x(2+√3)-y(2+√3)}{2√3}\)(\(√3\) - 2)

On Simplifying, we get m = \(\frac{x-y}{2√3}\)

Now, Consider triangle MOA,

Applying Pythagoras theorem(Since Triangle is right-angled at O)

i.e. (OM)² + (OA)² = (MA)²

 m² + a² = (MA)²

(MA)² = ( \(\frac{x-y}{2√3}\))² +  (\(\frac{2(x-y)-( √3)(x+y)}{2√3}\))²

(MA)² = \(\frac{(x-y)^2+(2(x-y)-( √3)(x+y))^2}{(2√3)^2}\)

On Simplifying, We get

(MA)² = \((\frac{\text{x−y}}{2 √{3}})^2{\left{2-\frac{√{3}(\text{x+y})}{\text{x−y}}\right}}\)

(MA) = \(\frac{\text{x−y}}{2 \sqrt{3}} \sqrt{1+\left{2−\frac{\sqrt{3}(\text{x+y})}{\text{x−y}}\right}^2}\)

Thus, Length of the tower = \(\frac{\text{x−y}}{2 \sqrt{3}} \sqrt{1+\left{2−\frac{\sqrt{3}(\text{x+y})}{\text{x−y}}\right}^2}\)​

Concept:

• cosec (2nπ + θ) = cosec θ
• cosec (-θ) = - cosec θ

Calculation:

Given cosec\(\left(−\frac{73 π}{3}\right)\) 

⇒ cosec \(\left(24 \pi−\frac{73 π}{3}\right)\)

⇒ cosec \(\left(\frac{(72 -73) π}{3}\right)\)

⇒ cosec \(\left(-\frac{ π}{3}\right)\)

⇒ - cosec \(\left(\frac{ π}{3}\right)\) = \(−\frac{2}{\sqrt{3}}\)

∴ The correct option is (2).

Concept:

cos C + cos D = 2 cos \(\left( C + D \over 2\right)\) cos \(\left( C - D \over 2\right)\) __(i)

Calculation:

Given, \(\cos \left(\frac{5 \pi}{17}\right)+\cos \left(\frac{7 \pi}{17}\right)+2 \cos \left(\frac{11 \pi}{17}\right) \cos \left(\frac{\pi}{17}\right)\)

Using identity (i), 

⇒ \(2\cos \left(\frac{5 \pi}{17}+ \frac{7 \pi}{17} \over 2\right)\cos \left(\frac{5 \pi}{17}-\frac{7 \pi}{17}\over 2\right)+2 \cos \left(\frac{11 \pi}{17}\right) \cos \left(\frac{\pi}{17}\right)\)

⇒\(2\cos \left( \frac{6 \pi}{17}\right)\cos \left(-\frac{\pi}{17}\right)+2 \cos \left(\frac{11 \pi}{17}\right) \cos \left(\frac{\pi}{17}\right)\)

⇒ \(2[\cos \left( \frac{6 \pi}{17}\right)\cos \left(\frac{\pi}{17}\right)+ \cos \left(\frac{11 \pi}{17}\right) \cos \left(\frac{\pi}{17}\right)]\)

⇒ \(2 \cos \left(\frac{\pi}{17}\right)[\cos \left( \frac{6 \pi}{17}\right)+ \cos \left(\frac{11 \pi}{17}\right)]\)

Again using (i),

⇒ \(2 \cos \left(\frac{\pi}{17}\right)[2\cos \left( \frac{6 \pi}{17}+ \frac{11 \pi}{17} \over 2\right)\cos \left(\frac{6 \pi}{17}-\frac{11 \pi}{17}\over 2\right)]\)

⇒ \(4 \cos \left(\frac{\pi}{17}\right)\cos \left( \frac{ \pi}{2}\right)\cos \left(\frac{5 \pi}{34}\right)\)

⇒ \(4 \cos \left(\frac{\pi}{17}\right)\times0\times \cos \left(\frac{5 \pi}{34}\right)\) = 0

∴ The correct option is (1).

Concept:

Use trigonometric identity:

\(tan2a=\frac{2tana}{1-tan^{2}a}\)

Solution:

Let \(tan\left ( \frac{3\pi }{8} \right )=t\)

\(a=\frac{3\pi }{8}\)

\(tan\left ( \frac{3\pi }{4} \right )=-1\)

\(tan\frac{3\pi }{4}=\frac{2tan\left ( \frac{3\pi }{8} \right )}{1-tan^{2}\left ( \frac{3\pi }{8} \right )}\)

⇒ \(-1=\frac{2t}{1-t^{2}}\)

⇒ 2t = t² - 1

⇒ t² - 2t - 1 = 0

⇒ \(t=\frac{2\pm 2\sqrt{2}}{2}\)

⇒ \(t=1\pm \sqrt{2}\)

⇒ \(t=1+\sqrt{2}\) (Quadrant I)

∴ The correct option is (2)

Concept:

1 + cot²x = cosec²x

Calculation:

let cosec−1 2 = y

As per the graph of cosec−1 x, cosec−1 2 will lie in (0, π/2]

⇒ cosec y = 2

⇒ cot²y = cosec²y - 1 = 2² - 1 = 3

⇒ cot y = √3

Given expression tan−1 cot(cosec−1 2) 

Putting the values in the expression,

⇒ tan−1 cot(cosec−1 2) = tan−1 cot(y) 

⇒ tan−1 cot(cosec−1 2) = tan−1 (√3) 

⇒ tan−1 cot(cosec−1 2) =  \(\frac{π}{3}\) which is in (0, π/2] so it is acceptable.

∴ The correct option is (4).

Concept:

For a triangle ABC, with sides a, b, c opposite to angle A, B, C respectively.

Cosine formula-

\(\cos A = {b^2 +c^2 -a^2 \over 2bc}\), \(\cos B = {a^2 +c^2 -b^2 \over 2ac}\) and \(\cos C = {a^2 +b^2 -c^2 \over 2ab}\)

• cos 3x = 4cos³x - 3cos x

Calculation:

Given: a triangle ABC with a = 4, b = 3, c = 2. 

⇒ \(\cos C = {a^2 +b^2 -c^2 \over 2ab} = {4^2 +3^2 -2^2 \over 2(4)(3)}\)

⇒ \(\cos C = {16 + 9 - 4 \over 24} = {21 \over 24}\)

⇒ cos C = \({7 \over 8}\)

So cos 3C = 4cos3C - 3cos C

⇒ \(\cos 3C = 4({7 \over 8})^3 - 3.{7 \over 8}\)

⇒ \(\cos 3C = 4({343 \over 512}) - {21\over 8}\)

⇒ \(\cos 3C = {343 \over 128} - {21\over 8}\)

⇒ \(\cos 3C = {343 - 336\over 128} = {7 \over 128}\)

∴ The correct option is (1).

Concept:

cos C - cos D = - 2 sin \(\left( C + D \over 2\right)\) sin \(\left( C - D \over 2\right)\)

Values used:

sin 54° = \(\sqrt 5+ 1 \over 4\)

sin 18° = \(\sqrt 5- 1 \over 4\)

Calculation:

Given, cos 36° − cos 72° 

Using formula, 

cos 36° − cos 72° =  - 2 sin \(\left( 36 + 72 \over 2\right)\)° sin \(\left( 36 - 72 \over 2\right)\)

⇒ cos 36° − cos 72° =  - 2 sin 54° sin (-18)°

⇒ cos 36° − cos 72° =  2 sin 54° sin 18°

⇒ cos 36° − cos 72° =  2 \(\sqrt 5+ 1 \over 4\) \(\sqrt 5- 1 \over 4\)

⇒ cos 36° − cos 72° = \(5 - 1 \over 8\) = \(\frac{1}{2}\)

∴ The correct option is (3).

Concept:

• tan x = \(\sin x \over \cos x\)
• sec x = \(1 \over \cos x\)
• 1 + tan²x = sec²x
• In the fourth quadrant, cos x and sec x are positive and sin x, tan x, cosec x and cot x are negative.

Calculation:

Given sec x = \(\frac{25}{24}\) and x lies in the fourth quadrant, 

⇒ tan²x = sec²x - 1 = \(({25 \over24})^2 - 1 = {625 - 576 \over 576} \)

⇒ tan2x = \(\frac{49}{576}\)

⇒ tan x = - \(\frac{7}{24}\)  . . .(i){As x lies in fourth quadrant}

And since, tan x = \(\sin x \over \cos x\) = sin x. sec x

⇒ sin x = \(\tan x \over \sec x\) = - \(\frac{7}{24}\) × \(\frac{24}{25}\) = - \(\frac{7}{25}\)  . . .(ii)

So, from (i) and (ii),

tan x + sin x = - \(\frac{7}{24}\) - \(\frac{7}{25}\) = \(-7({ 25 + 24 \over 600}) = { - 7 \times 49 \over 600}\)

⇒ tan x + sin x = \(−\frac{343}{600}\)

∴ The correct option is (2).

Concept:

• tan(180° - θ) = -tan θ
• cot(180° - θ) = -cot θ
• tan 15° = 2 - \(\sqrt 3\)
• cot 15° = 2 + \(\sqrt 3\)

Formula used:

(a + b)² = a² + b² + 2ab

(a - b)2 = a2 + b2 - 2ab

Calculation:

Given,  tan2 165° + cot2 165° 

⇒  (tan (180 - 15)°)² + (cot (180 - 15)°)2 

⇒  (-tan2 15°) + (-cot2 15°)

⇒  tan2 15° + cot2 15° 

⇒  (2 - \(\sqrt 3\))2 + (2 + \(\sqrt 3\))2 

⇒ ( 4 + 3 - 4\(\sqrt 3\)) + (4 + 3 + 4\(\sqrt 3\))

⇒ tan2 165° + cot2 165° = 14

∴ The correct option is (2).

Concept:

• sin (2nπ + θ) = sin θ
• sin (-θ) = - sin θ
• sin (π - θ) = sin θ

Calculation:

Given,

\(\sin\left(2\text{n}π+\frac{5π}{6}\right)\sin\left(2\text{n}π−\frac{5π}{6}\right)\)

⇒ \(\sin\left(\frac{5π}{6}\right)\sin\left(−\frac{5π}{6}\right)\)

⇒ \(-\sin\left(\frac{5π}{6}\right)\sin\left(\frac{5π}{6}\right)\)

⇒ \(-\sin^2\left(\frac{5π}{6}\right)\)

⇒ \(-\sin^2\left(\pi - \frac{π}{6}\right)\)

⇒ \(-\sin^2\left(\frac{π}{6}\right) = -({1 \over 2})^2 \)

⇒ \(\sin\left(2\text{n}π+\frac{5π}{6}\right)\sin\left(2\text{n}π−\frac{5π}{6}\right)\) = \(−\frac{1}{4}\)

∴ The correct option is (1).

Concept:

• cos 2x = cos²x - sin²x
• If cos x = cos θ, then x = 2nπ ± θ (n ∈ Z)

Calculation:

If 1 + 2(sin x + cos x)(sin x − cos x) = 0

⇒ 1 + 2(sin² x - cos² x) = 0

⇒ 1 + 2(- cos 2x) = 0

⇒  2cos 2x = 1

⇒  cos 2x = \({1 \over 2}\), 

⇒  cos 2x = cos \(\pi \over 3\)

⇒ 2x = 2nπ ± \(\pi \over 3\)

⇒ x = nπ ± \(\pi \over 6\) , where 0 < x < 360°

when n = 0, x = \(\pi \over 6\) (only positive as 0 < x < 360°)

when n = 1, x = \(\pi + {\pi \over 6}, \pi - {\pi \over 6} = {7\pi \over 6}, {5\pi \over 6}\)

when n = 2, x = \(2\pi + {\pi \over 6}, 2\pi - {\pi \over 6} = {13\pi \over 6}, {11\pi \over 6}\)

where \(13 \pi \over 6\) is not in the range of x

Solutions are \(\pi \over 6\), \(5\pi \over 6\), \(7\pi \over 6\), \(11\pi \over 6\)

So when 0 < x < 360°, x has 4 solutions.

∴ The correct option is (4).

Concept:

The equation of the line passing through (h,k) and making an angle θ with the positive direction of the x-axis is (y - k) = tan θ (x - a) where tan θ is the slope of the line.

Value used:

• tan 75° = \(\sqrt 3 + 1 \over \sqrt 3 - 1\)

Calculation:

Given, the line passing through origin and inclining at an angle of 75° with the positive direction of x-axis.

1. The line passes through the point \(\left(1, \frac{1}{2−\sqrt{3}}\right)\).

The equation of the line passing through the origin (0,0) and inclining at an angle of 75° with the positive direction of the x-axis is (y - 0) = tan 75°(x - 0)

⇒ y = \(\sqrt 3 + 1 \over \sqrt 3 - 1\) x

⇒ \((\sqrt3 +1)x - (\sqrt 3 - 1) y =0\)

If line passes through a point then the point satisfies the equation,

For \(\left(1, \frac{1}{2−\sqrt{3}}\right)\),

 \((\sqrt3 +1)(1) - (\sqrt 3 - 1) ({1 \over 2 - \sqrt 3}) = {(\sqrt3 +1) (2 - \sqrt 3) - \sqrt 3 + 1\over 2 - \sqrt 3}\)

⇒ \({2\sqrt3 +2 - 3 -\sqrt 3- \sqrt 3 + 1\over 2 - \sqrt 3}\) = 0

⇒ \(\left(1, \frac{1}{2−\sqrt{3}}\right)\)lies on the line \((\sqrt3 +1)x - (\sqrt 3 - 1) y =0\)

So statement 1 is true.

2. The line entirely lies in the first and third quadrants.

The line passes through the origin (0,0) and inclining at an angle of 75° with the positive direction of the x-axis

Plotting the line on the graph from this information

It can clearly be seen that line only passes through first and third quadrant.

So, statement 2 is true.

∴ The correct option is (3).

Concept:

• Midpoint of (x1,y1) and (x2,y2) =  \(\left({x_1+ x_2 \over 2} , {y_1 + y_2 \over 2} \right)\)
• The equation of the line passing through (x₁,y₁) and (x₂,y₂) is

\((y - y_1) = {y_2- y_1 \over x_2 - x_1} (x - x_1)\)

Calculation:

Let the points of line on axes be (h,0) and (0,k) respectively

So, P(3, 4) is the mid-point of (h,0) and (0,k)

⇒ (3,4) = \(\left({h+ 0 \over 2} , {0+k \over 2} \right)\)

⇒ h = 6 and k = 8

So, the line passes through the points (6,0) and (0,8)

⇒ The equation of the line passing through (6,0) and (0,8) is

\((y - 0) = {8- 0 \over 0 -6} (x - 6)\)

⇒ -6y = 8x - 48

⇒ 4x + 3y − 24 = 0

∴ The correct option is (2).

Concept:

In an equilateral triangle, all sides are equal and each angle = 60° 

The slope of two parallel lines are equal

The slope of a line inclined to x-axis at an angle θ in positive direction = tan θ 

The equation of the line with slope m and passing through (a,b) is y - b = m(x - a)

• tan (180° + x) = tan x

Calculation:

Given: The base AB of an equilateral triangle ABC with side 8 cm lies along the y-axis such that the mid-point of AB is at the origin and B lies above the origin.

 

 From figure, it can be seen that AC makes an angle of (180 + 30) = 210° 

⇒ slope of AC = tan 210° = tan (180 + 30) = tan 30° = \(1 \over \sqrt 3\)

⇒ slope of the line parallel to AC = \(1 \over \sqrt 3\)

⇒ The equation of line passing through (8, 0) and parallel to side AC is

 (y - 0) = \(1 \over \sqrt 3\)(x - 8)

⇒ \(x -\sqrt 3 y - 8 = 0\)

∴ The correct option is (1).

Concept:

The equation of a circle with centre (a,b) and radius r is 

(x - a)2 + (y - b)2 = r2 

The points through which the circle passes satisfy this equation.

Calculation: 

Let the equation of the circle be 

(x - a)² + (y - b)² = r² __(1)

As the circle passes through the origin,

⇒ (0 - a)2 + (0 - b)2 = r2 

⇒  r2 = a2 + b2 

Putting it in (1),

equation of circle becomes (x - a)2 + (y - b)2 = a2 + b2 __(2)

As the circle is making positive intercepts 4 and 6 on the coordinate axes,

⇒ the circle passes through (4,0) and (0,6)

Put (4,0) in (2)

⇒ (4 - a)2 + (0 - b)2 = a2 + b2

⇒ 16 - 8a + a2 + b2 = a2 + b2

⇒ 16 = 8a or a = 2

And putting (0,6) in (2),

⇒ (0 - a)2 + (6 - b)2 = a2 + b2

⇒ a2 + 36 - 12b + b2 = a2 + b2

⇒ 36 = 12b or b = 3

So center of circle = (2,3)

From options, it passes through the line 3x − 4y + 6 = 0

∴ The correct option is (3).

Concept:

The equation of ellipse with centre (0,0) is of the form, \({x^2 \over a^2 } +{y^2 \over b^2} = 1\)

For an ellipse \({x^2 \over a^2 } +{y^2 \over b^2} = 1\) with y-axis as major axis, the eccentricity = \(e = \sqrt{1 - {a^2 \over b^2}}\)

Calculation:

Given: The centre of an ellipse is at (0, 0), major axis is on the y-axis and the ellipse passes through (3, 2) and (1, 6).

Let the equation of the ellipse be \({x^2 \over a^2 } +{y^2 \over b^2} = 1\)

(3,2) and (1,6) lies on the ellipse

⇒ \({3^2 \over a^2 } +{2^2 \over b^2} = 1\) and \({1^2 \over a^2 } +{6^2 \over b^2} = 1\)

⇒ \({9 \over a^2 } +{4 \over b^2} = 1\) __(i)

and \({1 \over a^2 } +{36 \over b^2} = 1\) __(ii)

Subtracting (i) from (ii),

⇒ \({1-9 \over a^2 } +{36 -4 \over b^2} = 0\)

 ⇒ 32a²​ = 8b²

⇒ 4a² = b²

Put in (i) 

⇒ \({9 \over a^2 } +{4 \over 4a^2} = 1\)

⇒ a² = 10

⇒ b² = 4 × 10 = 40 

So, the equation of the ellipse will be \({x^2 \over 10} +{y^2 \over 40} = 1\)

Now for the ellipse \({x^2 \over 10} +{y^2 \over 40} = 1\), the eccentricity of the ellipse is given by,

 \(e = \sqrt{1 - {10 \over 40}}\)

⇒ \(e = \sqrt{1 - {1 \over 4}}= \sqrt{ {3 \over 4}}\)

⇒ Eccentricity = \(\frac{\sqrt{3}}{2}\)

∴ The correct option is (1).