**Lines and Angles Questions for SSC CGL PDF**

Download SSC CGL Lines and Angles Questions with answersÂ PDF based on previous papers very useful for SSC CGL exams. Very Important Lines and Angles Questions for SSC exams

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**Question 1:Â **BC is the centre of the circle with centre O. A is a point on major arc BC as shown in the above figure. What is the value of $\angle{BAC}+\angle{OBC}$ ?

a)Â 120$^{\circ}$

b)Â 60$^{\circ}$

c)Â 90$^{\circ}$

d)Â 180$^{\circ}$

**Question 2:Â **AC and BC are two equal cords of a circle. BA is produced to any point P and CP, when joined cuts the circle at T. Then

a)Â CT : TP = AB : CA

b)Â CT : TP = CA : AB

c)Â CT : CB = CA : CP

d)Â CT : CB = CP : CA

**Question 3:Â **If an obtuse-angled triangle ABC, is the obtuse angle and O is the orthocenter. If = 54$^{\circ}$ , then is

a)Â 108$^{\circ}$

b)Â 126$^{\circ}$

c)Â 136$^{\circ}$

d)Â 116$^{\circ}$

**Question 4:Â **The external bisectors of and of meet at point P. If = 80$^{\circ}$ , the is

a)Â 50$^{\circ}$

b)Â 40$^{\circ}$

c)Â 80$^{\circ}$

d)Â 100$^{\circ}$

**Question 5:Â **In a triangle ABC,Â $\angle{A} = 90^o , \angle{C} = 55^o , \overline{AD}$ is perpendicular to $\overline{BC}$. What is the value of $\angle{BAD}$ ?

a)Â 60 $^{\circ}$

b)Â 45 $^{\circ}$

c)Â 55 $^{\circ}$

d)Â 35 $^{\circ}$

**Question 6:Â **If O be the circumcentre of a triangle PQR and $\angle{QOR} = 110^o, \angle{OPR} = 25^o$, then the measure of $\angle{PRQ}$ is

a)Â 50 $^{\circ}$

b)Â 55 $^{\circ}$

c)Â 60 $^{\circ}$

d)Â 65 $^{\circ}$

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**Question 7:Â **A, B, C, D are four points on a circle. AC and BD intersect at a point such that $\angle{BEC} = 130^o$ and $\angle{ECD} = 20^o$. Then, $\angle{BAC}$ is

a)Â 90 $^{\circ}$

b)Â 100 $^{\circ}$

c)Â 110 $^{\circ}$

d)Â 120 $^{\circ}$

**Question 8:Â **ABC is a triangle. The bisectors of the internal angle $\angle$B and external $\angle$C intersect at D. If $\angle$BDC=$50^{\circ}$, then $\angle$A is

a)Â $100^{\circ}$

b)Â $90^{\circ}$

c)Â $120^{\circ}$

d)Â $60^{\circ}$

**Question 9:Â **In a triangle ABC, the side BC is extended up to D. Such that CD = AC, if angleBAD = $109Â°$ and angleACB=$72Â°$ then the value of angleABC is

a)Â $35Â°$

b)Â $60Â°$

c)Â $40Â°$

d)Â $45Â°$

**Question 10:Â **The line passing through (-2,5) and (6,b) is perpendicular to the line 20x + 5y = 3. Find b?

a)Â -7

b)Â 4

c)Â 7

d)Â -4

**Question 11:Â **The co-ordinates of the centroid of a triangle ABC are (-1,-2) what are the co-ordinates of vertex C, if co-ordinates of A and B are (6,-4)Â and (-2,2) respectively?

a)Â (-7,-4)

b)Â (7,4)

c)Â (7,-4)

d)Â (-7,4)

**Question 12:Â **Find equation of the perpendicular bisector of segment joining the points (2,-6) and (4,0)?

a)Â x + 3y = 6

b)Â x + 3y = -6

c)Â x – 3y = -6

d)Â x – 3y = 6

**Question 13:Â ** Â In what ratio is the segment joining (12,Â1) and (Â3,4) divided by the Y Âaxis?

a)Â 4:1

b)Â 1:4

c)Â 4:3

d)Â 3:4

**Question 14:Â **The line passing through (4,3) and (y,0) is parallel to the line passing through (Â1,2) and (3,0). Find y?

a)Â Â1

b)Â 7

c)Â 2

d)Â Â5

**Question 15:Â **What is the slope of the line perpendicular to the line passing through the points (8,Â2) and (3,Â1)?

a)Â -5

b)Â 3/5

c)Â 5/3

d)Â 1/5

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**Answers & Solutions:**

**1)Â AnswerÂ (C)**

As we know that angle through an arc on centre is double that of made on remaining arc.

HenceÂ $\angle{BOC}=2\angle{BAC} $=2x (where x=$\angle{BAC} $)

andÂ Â $\angle{OBC}$ would be 90-x

SoÂ Â $\angle{BAC}+\angle{OBC}$=90

**2)Â AnswerÂ (C)**

It is given that AC = BC, also $\triangle$ PTB andÂ $\triangle$Â PAC are similar, we haveÂ :

$\frac{CA}{CP}=\frac{BT}{BP}$ —————-(i)

Also, we haveÂ $\angle$Â PBC =Â $\angle$Â BTC ($\because$Â $\angle$Â PBC = $\angle$ BAC = $\angle$Â BTC) andÂ $\angle$Â PCB =Â $\angle$Â BCT

=>Â $\triangle$Â PBC $\sim$ $\triangle$Â BTC

Thus,Â $\frac{CB}{BP}=\frac{CT}{BT}$

=>Â $\frac{BT}{BP}=\frac{CT}{CB}$ ————–(ii)

From equations (i) and (ii), we getÂ :

$\frac{CA}{CP}=\frac{CT}{CB}$

=> Ans – (C)

**3)Â AnswerÂ (B)**

**4)Â AnswerÂ (A)**

**5)Â AnswerÂ (C)**

$\angle A = 90^o$

$\angle C = 55^o$

$\angle B$ will be $180 – (90+55) = 35^o$

As AD is perpendicular to BC Hence $\angle BAD=180-(90+35)=55^o$

**6)Â AnswerÂ (C)**

As we know circumcentre O is perpendicular bisector of sides of a triangle.

And $\angle QOR = 110^o$

and OQ=OR (radius) hence angles OQR and ORQ will be also be equal.

Which will have value equal to $\frac{180-110}{2} = 35^o$

Now angle OPR and PRO will also have equal value as $25^o$.

So angle PQR will be 35+25 = $60^o$

**7)Â AnswerÂ (C)**

Angle ABD will be equal to angle ACD = $20^o$ (same sector angles)

Angle BEC = $130^o$ so angle AED = $130^o$ (concurrent angles)

Now angle BEA will be $\frac{360-130-130}{2} = 50^o$

So angle EDC will be $180-(50+20) = 110^o$

**8)Â AnswerÂ (A)**

In $\triangle$ BDC,

=> $y+(180^\circ-2x+x)+50^\circ=180^\circ$

=> $y-x+50^\circ=0$

=> $y-x=-50^\circ$

In $\triangle$ ABC,

=> $2y+(180^\circ-2x)+\angle A=180^\circ$

=> $2(y-x)+\angle A=0$

=> $2(-50^\circ)+\angle A=0$

=> $\angle A=100^\circ$

=> Ans – (A)

**9)Â AnswerÂ (A)**

Angle ACB =72$^o$

Angle ACD = 108$^o$

Angle CAD = Angle ADC = 36$^o$

Angle BAD = Angle BAC + Angle CAD =109$^o$

Angle BAC = 73$^o$

Angle ABC = 180 – Angle BAC – Angle ACB = 180 – 73 – 72 = 35$^o$

Hence Option A is the correct answer.

**10)Â AnswerÂ (C)**

Slope of line having equation : $ax + by + c = 0$ is $\frac{-a}{b}$

=> Slope of line $20x + 5y = 3$ is $\frac{-20}{5} = -4$

SlopeÂ line passing through (-2,5) and (6,b) = $\frac{b – 5}{6 + 2} = \frac{(b – 5)}{8}$

Also, product of slopes of two perpendicular lines is -1

=> $\frac{(b – 5)}{8} \times -4 = -1$

=> $b – 5 = \frac{8}{4} = 2$

=> $b = 2 + 5 = 7$

=> Ans – (C)

**11)Â AnswerÂ (A)**

Coordinates of centroid of triangle with vertices $(x_1 , y_1)$ , $(x_2 , y_2)$ and $(x_3 , y_3)$ is $(\frac{x_1 + x_2 + x_3}{3} , \frac{y_1 + y_2 + y_3}{3})$

Let coordinates of vertex C = $(x , y)$

Vertex A(6,-4) and Vertex B(-2,2) and Centroid = (-1,-2)

=> $-1 = \frac{-2 + 6 + x}{3}$

=> $x + 4 = -1 \times 3 = -3$

=> $x = -3 – 4 = -7$

Similarly, => $-2 = \frac{-4 + 2 + y}{3}$

=> $y – 2 = -2 \times 3 = -6$

=> $y = -6 + 2 = -4$

$\therefore$ Coordinates of vertex C = (-7,-4)

=> Ans – (A)

**12)Â AnswerÂ (B)**

Let line $l$ perpendicularly bisects line joining Â A(2,-6) and B(4,0) at C, thus C is the mid point of AB.

=> Coordinates of C = $(\frac{2 + 4}{2} , \frac{-6 + 0}{2})$

= $(\frac{6}{2} , \frac{-6}{2}) = (3,-3)$

Now, slope of AB = $\frac{y_2 – y_1}{x_2 – x_1} = \frac{(0 + 6)}{(4 – 2)}$

= $\frac{6}{2} = 3$

Let slope of line $l = m$

Product of slopes of two perpendicular lines = -1

=> $m \times 3 = -1$

=> $m = \frac{-1}{3}$

Equation of a line passing through point $(x_1,y_1)$ and having slope $m$ is $(y – y_1) = m(x – x_1)$

$\therefore$ Equation of line $l$

=> $(y + 3) = \frac{-1}{3}(x – 3)$

=> $3y + 9 = -x + 3$

=> $x + 3y = 3 – 9 = -6$

=> Ans – (B)

**13)Â AnswerÂ (A)**

Using section formula, the coordinates of point that divides line joining A = $(x_1 , y_1)$ and B = $(x_2 , y_2)$ in the ratio aÂ : b

= $(\frac{a x_2 + b x_1}{a + b} , \frac{a y_2 + b y_1}{a + b})$

Let the ratio in whichÂ the segment joining (12,1) and (3,4) divided by the y-axis = $k$ : $1$

Since, the line segment is divided by y-axis, thus x coordinate of the point will be zero, let the point of intersection = $(0,y)$

Now, point P (0,y) divides (12,1) and (3,4) in ratio = k : 1

=> $0 = \frac{(3 \times k) + (12 \times 1)}{k + 1}$

=> $3k + 12 = 0$

=> $k = \frac{-12}{3} = -4$

$\therefore$ Line segment joining (12,Â1) and (Â3,4) is divided by the Y Âaxis in the ratio = 4 : 1 externally

=> Ans – (A)

**14)Â AnswerÂ (B)**

Slope of line passing through $(x_1,y_1)$ and $(x_2,y_2)$ is $\frac{y_2 – y_1}{x_2 – x_1}$

=>Â Slope of line passing throughÂ (1,2) and (3,0) = $\frac{0 – 2}{3 – 1} = \frac{-2}{2} = -1$

Slope of line passing throughÂ (4,3) and (y,0) = $\frac{0 – 3}{y – 4} = \frac{-3}{(y – 4)}$

Also, slopes of parallel lines are equal.

=> $\frac{-3}{y – 4} = -1$

=> $y – 4 = 3$

=> $y = 3 + 4 = 7$

=> Ans – (B)

**15)Â AnswerÂ (A)**

Slope of line passing through $(x_1,y_1)$ and $(x_2,y_2)$ is $\frac{y_2 – y_1}{x_2 – x_1}$

=>Â Slope of line passing throughÂ (3,1) and (8,2) = $\frac{2 – 1}{8 – 3} = \frac{1}{5}$

Let slope of line perpendicular to it = $m$

Also, product of slopes of two perpendicular lines = -1

=> $m \times \frac{1}{5} = -1$

=> $m = -5$

=> Ans – (A)

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