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 $\mathrm{\angle }BAC+\mathrm{\angle }OBC$ ?

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, $\mathrm{\angle }A={90}^o,\mathrm{\angle }C={55}^o,\overline{AD}$ is perpendicular to $\overline{BC}$. What is the value of $\mathrm{\angle }BAD$ ?

a) 60 ${}^{\circ }$

b) 45 ${}^{\circ }$

c) 55 ${}^{\circ }$

d) 35 ${}^{\circ }$

Question 6: If O be the circumcentre of a triangle PQR and $\mathrm{\angle }QOR={110}^o,\mathrm{\angle }OPR={25}^o$, then the measure of $\mathrm{\angle }PRQ$ 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 $\mathrm{\angle }BEC={130}^o$ and $\mathrm{\angle }ECD={20}^o$. Then, $\mathrm{\angle }BAC$ 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 $\mathrm{\angle }$B and external $\mathrm{\angle }$C intersect at D. If $\mathrm{\angle }$BDC=${50}^{\circ }$, then $\mathrm{\angle }$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 $\mathrm{\angle }BOC=2\mathrm{\angle }BAC$=2x (where x=$\mathrm{\angle }BAC$)
and  $\mathrm{\angle }OBC$ would be 90-x
So  $\mathrm{\angle }BAC+\mathrm{\angle }OBC$=90

2) Answer (C)

It is given that AC = BC, also $\mathrm{△}$ PTB and $\mathrm{△}$ PAC are similar, we have :

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

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

=> $\mathrm{△}$ PBC $\sim$ $\mathrm{△}$ 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)

$\mathrm{\angle }A={90}^o$
$\mathrm{\angle }C={55}^o$
$\mathrm{\angle }B$ will be $180–(90+55)={35}^o$
As AD is perpendicular to BC Hence $\mathrm{\angle }BAD=180-(90+35)={55}^o$

6) Answer (C)

As we know circumcentre O is perpendicular bisector of sides of a triangle.
And $\mathrm{\angle }QOR={110}^o$
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 $\mathrm{△}$ BDC,

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

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

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

In $\mathrm{△}$ ABC,

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

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

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

=> $\mathrm{\angle }A={100}^{\circ }$

=> 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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