Lines and Angles Questions for SSC CGL PDF

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Question 1: 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) 120$^{\circ}$

b) 60$^{\circ}$

c) 90$^{\circ}$

d) 180$^{\circ}$

Question 2: 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) CT : TP = AB : CA

b) CT : TP = CA : AB

c) CT : CB = CA : CP

d) CT : CB = CP : CA

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

a) 108$^{\circ}$

b) 126$^{\circ}$

c) 136$^{\circ}$

d) 116$^{\circ}$

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

a) 50$^{\circ}$

b) 40$^{\circ}$

c) 80$^{\circ}$

d) 100$^{\circ}$

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

a) 60 $^{\circ}$

b) 45 $^{\circ}$

c) 55 $^{\circ}$

d) 35 $^{\circ}$

Question 6: 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) 50 $^{\circ}$

b) 55 $^{\circ}$

c) 60 $^{\circ}$

d) 65 $^{\circ}$

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Question 7: 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) 90 $^{\circ}$

b) 100 $^{\circ}$

c) 110 $^{\circ}$

d) 120 $^{\circ}$

Question 8: 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) $100^{\circ}$

b) $90^{\circ}$

c) $120^{\circ}$

d) $60^{\circ}$

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

a) $35°$

b) $60°$

c) $40°$

d) $45°$

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

a) -7

b) 4

c) 7

d) -4

Question 11: 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)  and (-2,2) respectively?

a) (-7,-4)

b) (7,4)

c) (7,-4)

d) (-7,4)

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

a) x + 3y = 6

b) x + 3y = -6

c) x – 3y = -6

d) x – 3y = 6

Question 13:  ­ In what ratio is the segment joining (12,­1) and (­3,4) divided by the Y ­axis?

a) 4:1

b) 1:4

c) 4:3

d) 3:4

Question 14: The line passing through (4,3) and (y,0) is parallel to the line passing through (­1,2) and (3,0). Find y?

a) ­1

b) 7

c) 2

d) ­5

Question 15: What is the slope of the line perpendicular to the line passing through the points (8,­2) and (3,­1)?

a) -5

b) 3/5

c) 5/3

d) 1/5

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Answers & Solutions:

1) Answer (C)

As we know that angle through an arc on centre is double that of made on remaining arc.
Hence $\angle{BOC}=2\angle{BAC} $=2x (where x=$\angle{BAC} $)
and  $\angle{OBC}$ would be 90-x
So  $\angle{BAC}+\angle{OBC}$=90

2) Answer (C)

It is given that AC = BC, also $\triangle$ PTB and $\triangle$ PAC are similar, we have :

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

Also, we have $\angle$ PBC = $\angle$ BTC ($\because$ $\angle$ PBC = $\angle$ BAC = $\angle$ BTC) and $\angle$ PCB = $\angle$ BCT

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

Thus, $\frac{CB}{BP}=\frac{CT}{BT}$

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

From equations (i) and (ii), we get :

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

=> Ans – (C)

3) Answer (B)

4) Answer (A)

5) Answer (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) Answer (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) Answer (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) Answer (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) Answer (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) Answer (C)

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

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

Slope 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) Answer (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) Answer (B)

Let line $l$ perpendicularly bisects line joining  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) Answer (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 a : b

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

Let the ratio in which 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 (12,­1) and (­3,4) is divided by the Y ­axis in the ratio = 4 : 1 externally

=> Ans – (A)

14) Answer (B)

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

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

Slope of line passing through (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) Answer (A)

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

=> Slope of line passing through (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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