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Maths M.C.Q

Lines and Angles · Gujarat Board (GSEB) STD 9 (GSEB - English Medium). 219 questions.

Questions · Page 2 of 5

M.C.Q

MCQ 521 Mark
Side BC of $\triangle\text{ABC}$ has been produced to D on left-hand side and to E on right-hand side such that $\angle\text{ABD}=125^\circ$ and $\angle\text{ACE}=130^\circ.$ Then $\angle\text{A}=?$
  • A
    50º
  • B
    75º
  • C
    55º
  • D
    65º
Answer
  1. 75º
    Solution:
    $\angle\text{ABD}+\angle\text{ABC}=180^\circ$ (Linear Pair)
    $\angle\text{ABC}=180^\circ-125^\circ=55^\circ$
    $\angle\text{ACE}+\angle\text{ACB}=180^\circ$ (Linear Pair)
    $\angle\text{ACB}=180^\circ-130^\circ=50^\circ$
    In $\triangle\text{ABC}$
    $\angle\text{ABC}+\angle\text{ACB}+\angle\text{BAC}=180^\circ$ (Angle sum property)
    $\angle\text{BAC}=180^\circ-50^\circ-55^\circ$
    $\angle\text{BAC}=75^\circ.$
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MCQ 531 Mark
An angle is one-fifth of its supplement. The measure of the angle is:
  • A
    75º
  • B
    15º
  • C
    30º
  • D
    15º
Answer
  1. 30º
    Solution:
    Let one angle be xº
    Its supplementary angle will be 180º - xº
    According to question
    $\text{x}=\frac{1}{5}(180^\circ-\text{x})$
    5x + x = 180º
    6x = 180º
    $\text{x}=\frac{180}{6}$
    x = 30º.
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MCQ 551 Mark
If two angles are complements of each other, then each angle is:
  • A
    An acute angle.
  • B
    An obtuse angle.
  • C
    A right angle.
  • D
    A reflex angle.
Answer
  1. An acute angle.
    Solution:
    If two angles are complements of each other, that is, the sum of their measures is 90º, then each angle is an acute angle.
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MCQ 561 Mark
The number of triangles that can be drawn having angles as 50°, 60° and 70° are:
  • A
    None of these
  • B
    Two
  • C
    Only one
  • D
    Infinite
Answer
  1. Infinite
    Solution:
    As we know similar triangles can be drawn for any given triangle.
    These similar triangles will have the same angles as the original triangle
    $(\text{i.e}\angle50^\circ,\angle60^\circ$ and $\angle70^\circ)$ and will be infinite in number.
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MCQ 591 Mark
An exterior angle of a triangle is 105º and its two interior opposite angles are equal. Each of these equal angles is:
  • A
    $37\frac{1}{2}^\circ$
  • B
    $52\frac{1}{2}^\circ$
  • C
    $75^\circ$
  • D
    $72\frac{1}{2}^\circ$
Answer
  1. $52\frac{1}{2}^\circ$
    Solution:
    Let one of interior angle be xº
    Sum of two opposite interior angles = Exterior angle
    $\therefore$ xº + xº = 105º [ $\therefore$ Exterior angle = 105º (given)]
    $\Rightarrow2\text{x}^\circ=105^\circ$
    $\therefore\text{x}^\circ=\frac{105^\circ}{2}$
    $\Rightarrow\text{x}^\circ=52\frac{1}{2}$
    Hence, each angle of a triangle is $52\frac{1}{2}^\circ.$
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MCQ 601 Mark
In the given figure, $\angle\text{OEB}=75^\circ,\angle\text{OBE}=55^\circ$ and $\angle\text{OCD}=100^\circ.$ Then $\angle\text{ODC}=?$
  • A
    25º
  • B
    30º
  • C
    35º
  • D
    20º
Answer
  1. 30º
    Solution:
    In $\triangle\text{OEB}$
    $\angle\text{OEB}+\angle\text{EBO}+\angle\text{BOE}=180^\circ$ (Angle sum property)
    $75^\circ+55^\circ+\angle\text{BOE}=180^\circ$
    $\angle\text{BOE}=50^\circ$
    $\angle\text{BOE}=\angle\text{COD}=50^\circ$ (Vertically opposite angle)
    In $\triangle\text{ODC}$
    $\angle\text{ODC}+\angle\text{DOC}+\angle\text{DCO}=180^\circ$
    $\angle\text{ODC}=180^\circ-100^\circ-50^\circ$
    $\angle\text{ODC}=30^\circ.$
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MCQ 621 Mark
In the given figure, AB || CD. If $\text{AOC}=30^\circ$ and $\angle\text{OAB}=100^\circ$ then $\angle\text{OCD}=?$
  • A
    150º
  • B
    80º
  • C
    130º
  • D
    100º
Answer
  1. 130º
    Solution:

    Draw OE || AB || CD
    Now, OE || AB and OA is the transversal.
    $\therefore\angle\text{OAB}+\angle\text{AOE}=180^\circ$ [Angles on the same side of a transversal line are supplementary]
    $\Rightarrow\angle\text{OAB}+\angle\text{AOC}+\angle\text{COE}=180^\circ$
    $\Rightarrow100^\circ+30^\circ+\angle\text{COE}=180^\circ$
    $\Rightarrow\angle\text{COE}=50^\circ$
    Also,
    OE || CD and OC is the transversal.
    $\therefore\angle\text{OCD}+\angle\text{COE}=180^\circ$ [Angles on the same side of a transversal line are supplementary]
    $\Rightarrow\angle\text{OCD}+50^\circ=180^\circ$
    $\Rightarrow\angle\text{OCD}=130^\circ.$
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MCQ 631 Mark
  • A
    70º
  • B
    40º
  • C
    100º
  • D
    30º
Answer
  1. 40º
    Solution:
    Given that,
    l ll m
    Let, l || m and transversal cuts them and
    $\angle1=70^\circ$
    $\angle3=20^\circ$
    $\angle4=30^\circ$
    $\angle1+\angle2=180^\circ$ (Interior angle)
    $\angle2=110^\circ\text{(i})$
    $\angle2=\angle5$ (Vertically opposite angle)
    $\angle5=110^\circ\text{(ii)}$
    $\angle5+\angle3+\angle4=180^\circ$(Sum of angles of a triangle is 180o)
    110º + x + 30º = 180º
    x = 40º.
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MCQ 641 Mark
  • A
    $90-\frac{\text{x}}{2}$
  • B
    90 + 2x
  • C
    90 + x
  • D
    90 - 2x
Answer
  1. $90-\frac{\text{x}}{2}$
    Solution:
    Given that,
    11 || 12 and 13 || 14
    Let,
    $\angle1=\text{x}$
    $\angle2=\text{y}$
    $\angle3=\text{y}$
    $\angle1=\angle4$ (Alternate angle)
    $\angle4=\text{x}$
    $\angle5=\angle2$ (Vertically opposite angle)
    $\angle6=\angle3$ (Vertically opposite angle)
    $\angle5=\angle6=\text{y}$
    Now,
    $\angle4+\angle5+\angle6=180^\circ$ (Consecutive interior angle)
    $\text{y}=90-\frac{\text{x}}{2}.$
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MCQ 671 Mark
In the given figure, AB || CD || EF, $\text{EA }\bot\text{ AB}$ and BDE is the transversal such that $\angle\text{DEF}=55^\circ,$ Then $\angle\text{AEB}=?$
  • A
    35º
  • B
    55º
  • C
    45º
  • D
    25º
Answer
  1. 35º
    Solution:
    $\text{EA }\bot\text{ AB}$
    $\angle\text{AEF}=90^\circ$
    $\angle\text{AEF}=\angle\text{BEF}+\angle\text{AEB}$
    $\angle\text{BEF}+\angle\text{AEB}=90^\circ$
    $\angle\text{BEF}=55^\circ$
    $55^\circ+\angle\text{AEB}=90^\circ$
    $\text{AEB}=90^\circ-55^\circ$
    $\angle\text{AEB}=35^\circ.$
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MCQ 681 Mark
If two lines intersect each other then:
  • A
    Corresponding angles are equal.
  • B
    Alternate interior angles are equal.
  • C
    Co-interior angles are equal.
  • D
    Vertically opposite angles are equal.
Answer
  1. Vertically opposite angles are equal.
    Solution:

    $\angle\text{A}+\angle\text{B}=180^\circ$ (Linear Pair)
    $\angle\text{B}+\angle\text{C}=180$ (Linear Pair)
    On equating above equations, we get
    $\angle\text{A}+\angle\text{B}=\angle\text{B}+\angle\text{C}$
    $\angle\text{A}=\angle\text{C}$
    Similarly, $\angle\text{B}=\angle\text{D}$
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MCQ 691 Mark
If one of the angles of a triangle is 130º, then the angle between the bisectors of the other two angles can be:
  • A
    145º
  • B
    155º
  • C
    50º
  • D
    65º
Answer
  1. 50º
    Solution:
    Let angles of a triangle be $\angle\text{A},\angle\text{B}$ and $\angle\text{C}.$

    In $\triangle\text{ABC},$
    $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ $[sum of all interior angles of a triangle is 180º]
    $\Rightarrow12\angle\text{A}+12\angle\text{B}+12\angle\text{C}=180^\circ2=90^\circ$[dividing both sides by 2]
    $\Rightarrow12\angle\text{B}+12\angle\text{C}=90^\circ−12\angle\text{A} $
    $[\because\text{In}\triangle\text{OBC},\angle\text{OBC}+\angle\text{BCO}+\angle\text{COB}=180^\circ]$
    ⇒ Since, $\angle\text{B}2+\angle\text{C}2+\angle\text{BOC}=180^\circ$ as BO and OC are the angle bisectors of $\angle\text{ABC} $ and (\angle\text{BCA},$ respectively
    $\Rightarrow180^\circ−\angle\text{BOC}=90^\circ−12\angle\text{A}$
    $\therefore\angle\text{BOC}=180^\circ-90^\circ+12\angle\text{A}$
    = 90º + 12 × 130º = 90º + 65º $[\therefore\angle\text{A}=130^\circ(\text{given)}]$
    = 155º
    Hence, the required angle is 155º.
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MCQ 731 Mark
  • A
    85º
  • B
    65º
  • C
    20º
  • D
    45º
Answer
  1. 45º
    Solution:
     l || m
    Let transversal be n and $\angle1=65^\circ$
    $\angle2=20^\circ$
    $\angle3=\text{x}$
    Since,
    l || m and n cuts them so,
    $\angle1+\angle4=180^\circ$ (Co. interior angle)
    $65^\circ+\angle4=80^\circ$
    $\angle4=115^\circ\text{(i)}$
    $\angle4=\angle5=115^\circ$ (Vertically opposite angle)
    $\angle2+\angle5+\angle3=180^\circ$
    20º + 115º + x = 180º
    x = 45º.
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MCQ 741 Mark
In the given figure, AB || CD. If $\angle\text{AOC}=30^\circ$ and $\angle\text{OAB}=100^\circ$ then $\angle\text{OCD}=?$
Answer
  1. 130º
    Solution:
    Construction: Through O, draw OE || AB || CD
    $\Rightarrow\angle\text{BAO}+\angle\text{EOA}=180^\circ$
    $\Rightarrow100^\circ+\angle\text{EOA}=180^\circ$
    $\Rightarrow\angle\text{EOA}=80^\circ$
    So, $\angle\text{EOC}=\angle\text{EAO}-\angle\text{COA}=80^\circ-30^\circ=50^\circ$
    Since CD || EO
    $\angle\text{OCD}+\angle\text{EOC}=1806^\circ$
    $\Rightarrow\angle\text{OCD}+50^\circ=180^\circ$
    $\Rightarrow\angle\text{OCD}=130^\circ$
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MCQ 751 Mark
In fig if x = 30º then y =
  • A
    210º
  • B
    180º
  • C
    90º
  • D
    150º
Answer
  1. 150º
    Solution:
    x + y = 180º (linear pair)
    x = 30º
    30º + y = 180º
    y = 180º - 30º
    y = 150º.
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MCQ 761 Mark
In the adjoining figure $\angle\text{QPR}=62^\circ$ and $\angle\text{PRQ}=64^\circ$ If OQ and OR and bisectors of $\angle\text{PQR}$ and $\angle\text{PRQ}$ respectively, then $\angle\text{OQR}$ and $\angle\text{QOR}:$
  • A
    27º, 121º
  • B
    20º, 80º
  • C
    26º, 124º
  • D
    121º, 20º
Answer
  1. 27º, 121º
    Solution:
    In $\triangle\text{PQR}$
    $\angle\text{QPR}+\angle\text{PQR}+\angle\text{PRQ}=180^\circ$ (Angle sum property)
    $\angle\text{PQR}=180^\circ-62^\circ-64^\circ$
    $\angle\text{PQR}=54^\circ$
    $\angle\text{ORQ}=32^\circ$ (OR is a bisector)
    $\angle\text{OQR}=27^\circ$ OQ is a bisector)
    In $\triangle\text{OQR}$
    $\angle\text{OQR}+\angle\text{ORQ}+\angle\text{QOR}=180^\circ$ (Angle sum property)
    $\angle\text{QOR}=180^\circ-32^\circ-27^\circ=121^\circ.$
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MCQ 781 Mark
An exterior angle of a triangle is 80º and the interior opposite angles are in the ratio 1 : 3. Measure of each inte4rior opposite angle is:
  • A
    30º, 60º
  • B
    20º, 60º
  • C
    30º, 90º
  • D
    40º, 120º
Answer
  1. 20º, 60º
    Solution:
    let the common ratio is x
    the ratio of interior angles are 1 : 3
    so angles are x and 3x
    x + 3x = 80
    $\text{x}=\frac{80}{4}$
    x = 20
    So angles are 20º and 60º
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MCQ 791 Mark
In the given figure, AB || CD. If $\angle\text{EAB}=50^\circ$ and $\angle\text{ECD}=60^\circ$ then $\angle\text{AEB}=?$
  • A
    50º
  • B
    60º
  • C
    70º
  • D
Answer
  1. 70º
    Solution:
    Let $\angle\text{AEB}=\text{x}^\circ$
    Now, AB || CD and BC is the transversal
    $\therefore\angle\text{ABE}=\angle\text{BCD}=60^\circ$ (Alterante angles)
    In $\triangle\text{ABE},$
    $\angle\text{BAE}+\angle\text{AEB}+\angle\text{ABE}=180^\circ$ (Angle sum property)
    $\Rightarrow50^\circ+\text{x}^\circ+60^\circ=180^\circ$
    $\Rightarrow\text{x}=70^\circ$
    $\therefore\angle\text{AEB}=70^\circ$
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MCQ 811 Mark
  • A
    100
  • B
    105
  • C
    110
  • D
Answer
  1. 100
    Solution:

    Extending line BA and CP to meet at point E.
    $\angle\text{APE}=180^\circ-148^\circ=32^\circ$
    $\angle\text{EAP}=180^\circ-132^\circ=48^\circ$
    $\angle\text{AEP}=\text{x}^\circ$ [(Correspondence angles) because AB || CD cut by transverse EC]
    Now, in $\triangle\text{APE}$
    $\angle\text{APE}+\angle\text{EAP}+\angle\text{AEP}=180^\circ$
    $\Rightarrow\ 32^\circ+48^\circ+\text{x}^\circ=180^\circ$
    $\Rightarrow\ \text{x}^\circ=100$
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MCQ 821 Mark
  • A
    72º
  • B
    60º
  • C
    48º
  • D
    80º
Answer
  1. 60º
    Solution:
    Let $\angle\text{AOC}=\text{x}^\circ=(4\text{a})^\circ,\angle\text{COD}=\text{y}^\circ=(5\text{a})^\circ$ and $\angle\text{BOD}=\text{z}^\circ=(6\text{a})^\circ$
    Then, we have
    $\angle\text{AOC}+\angle\text{COD}+\angle\text{BOD}=180^\circ$ [Since AOB is a straight line]
    ⇒ 4a + 5a + 6a= 180º
    ⇒ 15a = 180º
    ⇒ a = 12º
    $\therefore$ y = 5 × a = 5 × 12º = 60º.
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MCQ 831 Mark
If two angles are supplementary and the larger is 20º less then three times the smaller, then the angles are:
  • A
    $72\frac{1}{2}^0,17\frac{1}{2}^0$
  • B
    $140^0,40^0$
  • C
    $130^0,50^0$
  • D
    $62\frac{1}{2}^0,27\frac{1}{2}^0$
Answer
  1. $130^0,50^0$
    Solution:
    Let the two supplimentary angles be x0 and 1800 - x0
    Let 1800 - x be the larger angle
    1800 - x = 3x - 200
    4x = 2000
    x = 500
    So the angles are 500 and 1300
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MCQ 851 Mark
  • A
    30º
  • B
    20º
  • C
    35º
  • D
    25º
Answer
  1. 20º
    Solution:
    Since, POQ is a line segment.
    $\therefore\angle\text{POQ}=180^\circ$
    $\Rightarrow\angle\text{POA}+\angle\text{AOB}+\angle\text{BOQ}=180^\circ$
    $\Rightarrow40^\circ+4\text{x}+3\text{x}=180^\circ$
    [Putting $\angle\text{POA}=40^\circ,\angle\text{AOB}=4\text{x}$ and $\angle\text{BOQ}=3\text{x}$]
    $\Rightarrow7\text{x}=180^\circ-40^\circ$
    $\Rightarrow7\text{x}=140^\circ$
    $\text{x}=\frac{140^\circ}{7}$
    $\text{x}=20^\circ.$
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MCQ 861 Mark
An angle is one fifth of its supplement. The measure of the angle is:
  • A
    15º
  • B
    30º
  • C
    75º
  • D
    150º
Answer
  1. 30º
    Solution:
    Let the measure of the angle be xº.
    So, its supplement = (180º - x)
    According to the given condition,
    $\text{x}=\frac{1}{5}(180^\circ-\text{x)}$
    $\Rightarrow5\text{x}=180-\text{x}$
    $\Rightarrow6\text{x}=180$
    $\Rightarrow\text{x}=30^\circ$
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MCQ 881 Mark
In the adjoining figure $\angle\text{QPR}=62^\circ$ and $\angle\text{PRQ}=64^\circ$ If OQ and OR and bisectors of $\angle\text{PQR}$ and $\angle\text{PRQ}$ respectively, then $\angle\text{OQR}$ and $\angle\text{QOR}:$
  • A
    121º, 20º
  • B
    27º, 121º
  • C
    20º, 80º
  • D
    26º, 124º
Answer
  1. 27º, 121º
    Solution:
    In $\triangle\text{PQR}$
    $\angle\text{QPR}+\angle\text{PQR}+\angle\text{PRQ}=180^\circ$ (Angle sum property)
    $\angle\text{PQR}=180^\circ-62^\circ-64^\circ$
    $\angle\text{PQR}=54^\circ$
    $\angle\text{ORQ}=32^\circ$ (OR is a bisector)
    $\angle\text{OQR}=27^\circ$ (OR is a bisector)
    In $\triangle\text{OQR}$
    $\angle\text{OQR}+\angle\text{ORQ}+\angle\text{QOR}=180^\circ$ (Angle sum property)
    $\angle\text{QOR}=180^\circ-32^\circ-27^\circ=121^\circ$
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MCQ 891 Mark
  • A
    12
  • B
    15
  • C
    18
  • D
Answer
  1. 20
    Solution:

    From figure,
    $\angle\text{ABD}+\angle\text{EBD}=180^\circ$
    $\Rightarrow\ \angle\text{EBD}=180^\circ-\angle\text{ABD}\dots(1)$
    Now,
    $\angle\text{ABD}=\text{y}^\circ+2\text{y}^\circ+\text{y}^\circ$
    $\Rightarrow\ \angle\text{ABD}=4\text{y}^\circ\dots(2)$
    Substituting (2) in (1), we have
    $\angle\text{EBD}=180^\circ-4\text{y}^\circ$
    Now,
    $\angle\text{EBD}=\angle\text{BDC}$ [Alternate angles]
    $\Rightarrow\ 180^\circ-4\text{y}^\circ=5\text{y}^\circ$
    $\Rightarrow\ 180^\circ=9\text{y}^\circ$
    $\Rightarrow\ \text{y}^\circ=20^\circ$
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MCQ 931 Mark
In the given figure, AOB is a straight line. If $\angle\text{AOC}=(3\text{x}-10)^\circ,\angle\text{COD}=50^\circ$ and $\angle\text{BOD}=(\text{x}+20)^\circ$ then $\angle\text{AOC}=?$
  • A
    40º
  • B
    60º
  • C
    80º
  • D
Answer
  1. 80º
    Solution:
    Since AOB is a straight line,
    $\angle\text{AOC}+\angle\text{COD}+\angle\text{BOD}=180^\circ$
    $\Rightarrow(3\text{x}-10)+50+(\text{x}+20)=180^\circ$
    $\Rightarrow4\text{x}+60=180$
    $\Rightarrow4\text{x}=120$
    $\Rightarrow\text{x}=30$
    So, $\angle\text{AOC}=3\text{x}-10=3(30)-10=80^\circ$
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MCQ 941 Mark
  • A
    20°
  • B
    30°
  • C
    45°
  • D
Answer
  1. 30°
    Solution:

    From figure, we can see
    $\angle\text{x}^\circ=\angle\text{y}^\circ$ [Vertically opposite angles]
    Also,
    $\angle\text{3}\text{x}^\circ+\angle\text{y}^\circ+\angle2\text{x}^\circ=180^\circ$
    Now,
    $\angle\text{x}^\circ=\angle\text{y}^\circ$
    $\therefore\ \angle\text{3}\text{y}^\circ+\angle\text{y}^\circ+\angle2\text{y}^\circ=180^\circ$
    $\Rightarrow\ \angle6\text{y}^\circ=180^\circ$
    $\Rightarrow\ \angle\text{y}^\circ=180^\circ$
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MCQ 961 Mark
In the given figure, AOB is a straight line. If $\angle\text{AOC}=(3\text{x} +10)^\circ$ and $\angle\text{BOC}=(4\text{x}-26)^\circ,$ then $\angle\text{BOC}=?$
Answer
  1. 86º
    Solution:
    Since AOB is a straight line,
    $\angle\text{AOC}+\angle\text{BOC}=180^\circ$
    $\Rightarrow(3\text{x}+10)+(4\text{x}-26)=180^\circ$
    $\Rightarrow7\text{x}-16=180^\circ$
    $\Rightarrow7\text{x}=196$
    $\Rightarrow\text{x}=28$
    So, $\angle\text{BOC}=4 \text{x}-26=4(28)-26=86^\circ$
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MCQ 981 Mark
Two straight lines AB and CD intersect one another at the point O. If $\angle\text{AOC}+\angle\text{COB}+\angle\text{BOD}=274^\circ,$ then $\angle\text{AOD}=$
  • A
    86°
  • B
    90°
  • C
    94°
  • D
    137°
Answer
  1. 86°
    Solution:

    $\angle\text{AOC}+\angle\text{COB}+\angle\text{BOD}+\angle\text{AOD}=360^\circ\dots(1)$
    Now,
    $\angle\text{AOC}+\angle\text{COB}+\angle\text{BOD}=274^\circ\dots(2)$ [Given]
    From (1) and (2).
    $274^\circ+\angle\text{AOD}=360^\circ$
    $\Rightarrow\ \angle\text{AOD}=360^\circ-274^\circ$
    $\Rightarrow\ \angle\text{AOD}=86^\circ$
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MCQ 991 Mark
Two complementary angles are such that twich the measure of the one is equal to three times the measure of the other. The larger of the two measure.
  • A
    72º
  • B
    54º
  • C
    63º
  • D
    36º
Answer
  1. 54º
    Solution:
    Let the measure of each angle be xº and (90 - x)º.
    According to the given condition,
    2x = 3(90 - x)
    ⇒ 2x = 270 - 3x
    ⇒ 5x = 270
    ⇒ x = 54º
    So, (90 - x)º = (90 - 54)º = 36º
    So, the larger of the two angles is 54º.
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MCQ 1001 Mark
In the given figure, $\angle\text{BAC}=40^\circ,\angle\text{ACB}=90^\circ$ and $\angle\text{BED}=100^\circ$ Then $\angle\text{BDE}=?$
  • A
    25°
  • B
    40°
  • C
    50°
  • D
    30°
Answer
  1. 30°
    Solution:
    In $\triangle\text{ABC}$
    $\angle\text{ABC}+\angle\text{BAC}+\angle\text{ACB}=180^\circ$(Angle sum property)
    $\angle\text{ABC}=180^\circ-90^\circ-40^\circ$
    $\angle\text{ABC}=50^\circ$
    In $\triangle\text{BED}$
    $\angle\text{BED}+\angle\text{EBD}+\angle\text{BDE}=180^\circ$(Angle sum property)
    $\angle\text{EBD}=180^\circ-50^\circ-100^\circ$
    $\angle\text{EBD}=30^\circ$
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