{"id":40053,"date":"2020-01-23T11:58:55","date_gmt":"2020-01-23T06:28:55","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=40053"},"modified":"2020-01-23T11:58:55","modified_gmt":"2020-01-23T06:28:55","slug":"lines-and-angles-questions-for-ssc-cgl-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/lines-and-angles-questions-for-ssc-cgl-pdf\/","title":{"rendered":"Lines and Angles Questions for SSC CGL PDF"},"content":{"rendered":"

Lines and Angles Questions for SSC CGL PDF<\/strong><\/span><\/h1>\n

Download SSC CGL Lines and Angles Questions with answers\u00a0 PDF based on previous papers very useful for SSC CGL exams. Very Important Lines and Angles Questions for SSC exams<\/p>\n

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Question 1:\u00a0<\/b>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 $\\angle{BAC}+\\angle{OBC}$ ?
\n\"\"<\/p>\n

a)\u00a0120$^{\\circ}$<\/p>\n

b)\u00a060$^{\\circ}$<\/p>\n

c)\u00a090$^{\\circ}$<\/p>\n

d)\u00a0180$^{\\circ}$<\/p>\n

Question 2:\u00a0<\/b>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<\/p>\n

a)\u00a0CT : TP = AB : CA<\/p>\n

b)\u00a0CT : TP = CA : AB<\/p>\n

c)\u00a0CT : CB = CA : CP<\/p>\n

d)\u00a0CT : CB = CP : CA<\/p>\n

Question 3:\u00a0<\/b>If an obtuse-angled triangle ABC, is the obtuse angle and O is the orthocenter. If = 54$^{\\circ}$ , then is<\/p>\n

a)\u00a0108$^{\\circ}$<\/p>\n

b)\u00a0126$^{\\circ}$<\/p>\n

c)\u00a0136$^{\\circ}$<\/p>\n

d)\u00a0116$^{\\circ}$<\/p>\n

Question 4:\u00a0<\/b>The external bisectors of and of meet at point P. If = 80$^{\\circ}$ , the is<\/p>\n

a)\u00a050$^{\\circ}$<\/p>\n

b)\u00a040$^{\\circ}$<\/p>\n

c)\u00a080$^{\\circ}$<\/p>\n

d)\u00a0100$^{\\circ}$<\/p>\n

Question 5:\u00a0<\/b>In a triangle ABC,\u00a0$\\angle{A} = 90^o , \\angle{C} = 55^o , \\overline{AD}$ is perpendicular to $\\overline{BC}$. What is the value of $\\angle{BAD}$ ?<\/p>\n

a)\u00a060 $^{\\circ}$<\/p>\n

b)\u00a045 $^{\\circ}$<\/p>\n

c)\u00a055 $^{\\circ}$<\/p>\n

d)\u00a035 $^{\\circ}$<\/p>\n

Question 6:\u00a0<\/b>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<\/p>\n

a)\u00a050 $^{\\circ}$<\/p>\n

b)\u00a055 $^{\\circ}$<\/p>\n

c)\u00a060 $^{\\circ}$<\/p>\n

d)\u00a065 $^{\\circ}$<\/p>\n

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Question 7:\u00a0<\/b>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<\/p>\n

a)\u00a090 $^{\\circ}$<\/p>\n

b)\u00a0100 $^{\\circ}$<\/p>\n

c)\u00a0110 $^{\\circ}$<\/p>\n

d)\u00a0120 $^{\\circ}$<\/p>\n

Question 8:\u00a0<\/b>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<\/p>\n

a)\u00a0$100^{\\circ}$<\/p>\n

b)\u00a0$90^{\\circ}$<\/p>\n

c)\u00a0$120^{\\circ}$<\/p>\n

d)\u00a0$60^{\\circ}$<\/p>\n

Question 9:\u00a0<\/b>In a triangle ABC, the side BC is extended up to D. Such that CD = AC, if angleBAD = $109\u00b0$ and angleACB=$72\u00b0$ then the value of angleABC is<\/p>\n

a)\u00a0$35\u00b0$<\/p>\n

b)\u00a0$60\u00b0$<\/p>\n

c)\u00a0$40\u00b0$<\/p>\n

d)\u00a0$45\u00b0$<\/p>\n

Question 10:\u00a0<\/b>The line passing through (-2,5) and (6,b) is perpendicular to the line 20x + 5y = 3. Find b?<\/p>\n

a)\u00a0-7<\/p>\n

b)\u00a04<\/p>\n

c)\u00a07<\/p>\n

d)\u00a0-4<\/p>\n

Question 11:\u00a0<\/b>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)\u00a0 and (-2,2) respectively?<\/p>\n

a)\u00a0(-7,-4)<\/p>\n

b)\u00a0(7,4)<\/p>\n

c)\u00a0(7,-4)<\/p>\n

d)\u00a0(-7,4)<\/p>\n

Question 12:\u00a0<\/b>Find equation of the perpendicular bisector of segment joining the points (2,-6) and (4,0)?<\/p>\n

a)\u00a0x + 3y = 6<\/p>\n

b)\u00a0x + 3y = -6<\/p>\n

c)\u00a0x – 3y = -6<\/p>\n

d)\u00a0x – 3y = 6<\/p>\n

Question 13:\u00a0<\/b> \u00ad In what ratio is the segment joining (12,\u00ad1) and (\u00ad3,4) divided by the Y \u00adaxis?<\/p>\n

a)\u00a04:1<\/p>\n

b)\u00a01:4<\/p>\n

c)\u00a04:3<\/p>\n

d)\u00a03:4<\/p>\n

Question 14:\u00a0<\/b>The line passing through (4,3) and (y,0) is parallel to the line passing through (\u00ad1,2) and (3,0). Find y?<\/p>\n

a)\u00a0\u00ad1<\/p>\n

b)\u00a07<\/p>\n

c)\u00a02<\/p>\n

d)\u00a0\u00ad5<\/p>\n

Question 15:\u00a0<\/b>What is the slope of the line perpendicular to the line passing through the points (8,\u00ad2) and (3,\u00ad1)?<\/p>\n

a)\u00a0-5<\/p>\n

b)\u00a03\/5<\/p>\n

c)\u00a05\/3<\/p>\n

d)\u00a01\/5<\/p>\n

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Answers & Solutions:<\/strong><\/span><\/p>\n

1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

As we know that angle through an arc on centre is double that of made on remaining arc.
\nHence\u00a0$\\angle{BOC}=2\\angle{BAC} $=2x (where x=$\\angle{BAC} $)
\nand\u00a0\u00a0$\\angle{OBC}$ would be 90-x
\nSo\u00a0\u00a0$\\angle{BAC}+\\angle{OBC}$=90<\/p>\n

2)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

<\/figure>\n

It is given that AC = BC, also $\\triangle$ PTB and\u00a0$\\triangle$\u00a0PAC are similar, we have\u00a0:<\/p>\n

$\\frac{CA}{CP}=\\frac{BT}{BP}$ —————-(i)<\/p>\n

Also, we have\u00a0$\\angle$\u00a0PBC =\u00a0$\\angle$\u00a0BTC ($\\because$\u00a0$\\angle$\u00a0PBC = $\\angle$ BAC = $\\angle$\u00a0BTC) and\u00a0$\\angle$\u00a0PCB =\u00a0$\\angle$\u00a0BCT<\/p>\n

=>\u00a0$\\triangle$\u00a0PBC $\\sim$ $\\triangle$\u00a0BTC<\/p>\n

Thus,\u00a0$\\frac{CB}{BP}=\\frac{CT}{BT}$<\/p>\n

=>\u00a0$\\frac{BT}{BP}=\\frac{CT}{CB}$ ————–(ii)<\/p>\n

From equations (i) and (ii), we get\u00a0:<\/p>\n

$\\frac{CA}{CP}=\\frac{CT}{CB}$<\/p>\n

=> Ans – (C)<\/p>\n

3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

5)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

$\\angle A = 90^o$
\n$\\angle C = 55^o$
\n$\\angle B$ will be $180 – (90+55) = 35^o$
\nAs AD is perpendicular to BC Hence $\\angle BAD=180-(90+35)=55^o$<\/p>\n

6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

As we know circumcentre O is perpendicular bisector of sides of a triangle.
\nAnd $\\angle QOR = 110^o$
\nand OQ=OR (radius) hence angles OQR and ORQ will be also be equal.
\nWhich will have value equal to $\\frac{180-110}{2} = 35^o$
\nNow angle OPR and PRO will also have equal value as $25^o$.
\nSo angle PQR will be 35+25 = $60^o$<\/p>\n

7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Angle ABD will be equal to angle ACD = $20^o$ (same sector angles)
\nAngle BEC = $130^o$ so angle AED = $130^o$ (concurrent angles)
\nNow angle BEA will be $\\frac{360-130-130}{2} = 50^o$
\nSo angle EDC will be $180-(50+20) = 110^o$<\/p>\n

8)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

<\/figure>\n

In $\\triangle$ BDC,<\/p>\n

=> $y+(180^\\circ-2x+x)+50^\\circ=180^\\circ$<\/p>\n

=> $y-x+50^\\circ=0$<\/p>\n

=> $y-x=-50^\\circ$<\/p>\n

In $\\triangle$ ABC,<\/p>\n

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

=> $2(y-x)+\\angle A=0$<\/p>\n

=> $2(-50^\\circ)+\\angle A=0$<\/p>\n

=> $\\angle A=100^\\circ$<\/p>\n

=> Ans – (A)<\/p>\n

9)\u00a0Answer\u00a0(A)<\/strong><\/p>\n


\nAngle ACB =72$^o$
\nAngle ACD = 108$^o$
\nAngle CAD = Angle ADC = 36$^o$
\nAngle BAD = Angle BAC + Angle CAD =109$^o$
\nAngle BAC = 73$^o$
\nAngle ABC = 180 – Angle BAC – Angle ACB = 180 – 73 – 72 = 35$^o$
\nHence Option A is the correct answer.<\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n

10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

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

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

Slope\u00a0line passing through (-2,5) and (6,b) = $\\frac{b – 5}{6 + 2} = \\frac{(b – 5)}{8}$<\/p>\n

Also, product of slopes of two perpendicular lines is -1<\/p>\n

=> $\\frac{(b – 5)}{8} \\times -4 = -1$<\/p>\n

=> $b – 5 = \\frac{8}{4} = 2$<\/p>\n

=> $b = 2 + 5 = 7$<\/p>\n

=> Ans – (C)<\/p>\n

11)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

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

Let coordinates of vertex C = $(x , y)$<\/p>\n

Vertex A(6,-4) and Vertex B(-2,2) and Centroid = (-1,-2)<\/p>\n

=> $-1 = \\frac{-2 + 6 + x}{3}$<\/p>\n

=> $x + 4 = -1 \\times 3 = -3$<\/p>\n

=> $x = -3 – 4 = -7$<\/p>\n

Similarly, => $-2 = \\frac{-4 + 2 + y}{3}$<\/p>\n

=> $y – 2 = -2 \\times 3 = -6$<\/p>\n

=> $y = -6 + 2 = -4$<\/p>\n

$\\therefore$ Coordinates of vertex C = (-7,-4)<\/p>\n

=> Ans – (A)<\/p>\n

12)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Let line $l$ perpendicularly bisects line joining \u00a0A(2,-6) and B(4,0) at C, thus C is the mid point of AB.<\/p>\n

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

= $(\\frac{6}{2} , \\frac{-6}{2}) = (3,-3)$<\/p>\n

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

= $\\frac{6}{2} = 3$<\/p>\n

Let slope of line $l = m$<\/p>\n

Product of slopes of two perpendicular lines = -1<\/p>\n

=> $m \\times 3 = -1$<\/p>\n

=> $m = \\frac{-1}{3}$<\/p>\n

Equation of a line passing through point $(x_1,y_1)$ and having slope $m$ is $(y – y_1) = m(x – x_1)$<\/p>\n

$\\therefore$ Equation of line $l$<\/p>\n

=> $(y + 3) = \\frac{-1}{3}(x – 3)$<\/p>\n

=> $3y + 9 = -x + 3$<\/p>\n

=> $x + 3y = 3 – 9 = -6$<\/p>\n

=> Ans – (B)<\/p>\n

13)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

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\u00a0: b<\/p>\n

= $(\\frac{a x_2 + b x_1}{a + b} , \\frac{a y_2 + b y_1}{a + b})$<\/p>\n

Let the ratio in which\u00a0the segment joining (12,1) and (3,4) divided by the y-axis = $k$ : $1$<\/p>\n

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

Now, point P (0,y) divides (12,1) and (3,4) in ratio = k : 1<\/p>\n

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

=> $3k + 12 = 0$<\/p>\n

=> $k = \\frac{-12}{3} = -4$<\/p>\n

$\\therefore$ Line segment joining (12,\u00ad1) and (\u00ad3,4) is divided by the Y \u00adaxis in the ratio = 4 : 1 externally<\/p>\n

=> Ans – (A)<\/p>\n

14)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

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

=>\u00a0Slope of line passing through\u00a0(1,2) and (3,0) = $\\frac{0 – 2}{3 – 1} = \\frac{-2}{2} = -1$<\/p>\n

Slope of line passing through\u00a0(4,3) and (y,0) = $\\frac{0 – 3}{y – 4} = \\frac{-3}{(y – 4)}$<\/p>\n

Also, slopes of parallel lines are equal.<\/p>\n

=> $\\frac{-3}{y – 4} = -1$<\/p>\n

=> $y – 4 = 3$<\/p>\n

=> $y = 3 + 4 = 7$<\/p>\n

=> Ans – (B)<\/p>\n

15)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

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

=>\u00a0Slope of line passing through\u00a0(3,1) and (8,2) = $\\frac{2 – 1}{8 – 3} = \\frac{1}{5}$<\/p>\n

Let slope of line perpendicular to it = $m$<\/p>\n

Also, product of slopes of two perpendicular lines = -1<\/p>\n

=> $m \\times \\frac{1}{5} = -1$<\/p>\n

=> $m = -5$<\/p>\n

=> Ans – (A)<\/p>\n

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We hope this Physics Questions for SSC CGL Exam preparation is so helpful to you.<\/p>\n","protected":false},"excerpt":{"rendered":"

Lines and Angles Questions for SSC CGL PDF Download SSC CGL Lines and Angles Questions with answers\u00a0 PDF based on previous papers very useful for SSC CGL exams. Very Important Lines and Angles Questions for SSC exams Take a free mock test for SSC CGL Download SSC CGL Previous Papers PDF More SSC CGL Important […]<\/p>\n","protected":false},"author":49,"featured_media":40057,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[3167,125,9,504,378,1493,1459,1611,1741,1441],"tags":[3465,462],"class_list":{"0":"post-40053","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-downloads-en","8":"category-featured","9":"category-ssc","10":"category-ssc-cgl","11":"category-ssc-chsl","12":"category-ssc-cpo","13":"category-ssc-gd","14":"category-ssc-je","15":"category-ssc-mts","16":"category-stenographer","17":"tag-lines-and-angles-questions","18":"tag-ssc-cgl"},"better_featured_image":{"id":40057,"alt_text":"Lines and Angles Questions for SSC CGL 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