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Lines And Angles Test - 6

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Lines And Angles Test - 6
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  • Question 1
    1 / -0
    In the given figure, find the value of ∠GOH if ∠BOC is 10° more than ∠AOH and ∠COD is 10° more than ∠BOC.

    Solution
    In the given figure,
    ∠EOF = 2x
    ∠AOB = ∠EOF = 2x (vertically opposite angles)
    Also, it is given that ∠BOC is 10° more than ∠AOH.
    i.e. ∠BOC = x + 10°
    And ∠COD is 10° more than ∠BOC.
    i.e. ∠COD = x + 10° + 10° = x + 20°
    Now, ∠AOH + ∠AOB + ∠BOC + ∠COD = 180° (because sum of all the angles on a straight line is 180°)
    x + 2x + x + 10° + x + 20° = 180°
    5x + 30° = 180°
    5x = 180° - 30°
    5x = 150°
    x =
    x = 30°
    ∠COD = ∠GOH = x + 20° = 30° + 20° = 50° (vertically opposite angles)
    Therefore, ∠GOH = 50°
  • Question 2
    1 / -0
    If the second angle of a triangle is 30° more than the first angle and the third angle is 30° less than the second angle, then what kind of triangle is this?
    Solution
    Let first angle = x
    According to the instructions given in the question,
    Second angle = x + 30°
    Third angle = x + 30° - 30° = x
    By angle sum property of triangles,
    x + x + 30° + x = 180°
    3x = 180° - 30° = 150°
    x = = 50°
    First angle = Third angle = x = 50°
    Since two angles of the triangle are equal, it is an isosceles triangle.
  • Question 3
    1 / -0
    In the given figure, MN ∥ PQ. Find the value of DCQ.

    Solution
    DAO = BCO = 40° (As MN || PQ, AC is the transversal and DAO and BCO are alternate interior angles)
    AOB = COD = 55° (Vertically opposite angles)
    In triangle COD, by angle sum property of triangles,
    COD + ODC + DCO = 180°
    55° + 60° + DCO = 180°
    DCO = 180° - 115° = 65°
    Now, BCO + DCO + DCQ = 180° (As sum of angles on a straight line is 180°)
    40° + 65° + DCQ = 180°
    DCQ = 180° - 105°
    DCQ = 75°
  • Question 4
    1 / -0
    In the given figure, AB ∥ CD ∥ EF, ∠1 = ∠2 + 50° and PQRS is a rhombus. Find the measure of ∠4.


    Solution
    It is given that
    ∠1 = ∠2 + 50° ... (i)
    Now, 70° + ∠1 + ∠2 = 180° (Co-interior angles)
    70° + ∠2 + 50° + ∠2 = 180° (From (i))
    2∠2 = 60°
    ∠2 = 30°
    ∠ARS = ∠RSQ = 70° (Alternate interior angles)
    ∠RSP = ∠RSQ + ∠QSP = 70° + 30° = 100°
    ∠RSP = ∠RQP (As PQRS is a rhombus and opposite angles of a rhombus are equal)
    100° = ∠2 + ∠3
    100° = 30° + ∠3
    ∠3 = 70°
    Now, ∠3 = ∠4 = 70°
  • Question 5
    1 / -0
    In the given figure, a = b + 25° and ∠COE = 90°. Find the measure of ∠BOC.

    Solution
    ∠AOB = ∠DOF (Vertically opposite angles)
    font face=Times new=serif=>a + b = 75°
    b + 25° + b = 75° (Given)
    2b = 75° - 25°
    2b = 50°
    b = 25°
    a = b + 25° = 25° + 25° = 50°
    a + ∠COD = 90° (Given)
    50° + ∠COD = 90°
    ∠COD = 40°
    ∠AOB + ∠BOC + ∠COD = 180° (Angles on a straight line)
    75° + ∠BOC + 40° = 180°
    ∠BOC = 180° - 115°
    ∠BOC = 65°
  • Question 6
    1 / -0
    In the given figure, PQ ∥ RS ∥ TU, and MN and KL are transversal lines. Also, ∠CED = ∠DEF and ∠EDF = 64°. What is the value of x?

    Solution

    In the given figure, ∠MAP = ∠ACR = ∠CET = 62° (Corresponding angles)
    ∠CEF = 180° - 62° (As sum of angles on a straight line is 180°)
    ∠CEF = 118°
    ∠CEF = ∠CED + ∠DEF
    118° = 2∠DEF [∠CED = ∠DEF (given)]
    ∠DEF = = 59°
    Now, in triangle EDF,
    59° + 64° + ∠EFD = 180° (Angle sum property of triangle)
    123° + ∠EFD = 180°
    ∠EFD = 180° - 123°
    ∠EFD = 57°
    ∠LFU = ∠EFD = 57° (Vertically opposite angles)
    ∠FDS = ∠EFD = 57° (Alternate interior angles)
    x + ∠FDS = 180° (Linear pair)
    x + 57° = 180°
    x = 180° - 57°
    x = 123°
  • Question 7
    1 / -0
    The measure of an angle is 45° less than the double of its supplementary angle. What is the measure of the angle?
    Solution
    Let the measure of the angle be x.
    Supplementary angle = 180° - x
    Now, according to the given instructions,
    x = 2(180° - x) - 45°
    x = 360° - 2x - 45°
    x + 2x = 360° - 45°
    3x = 315°
    x == 105°
    x = 105°
  • Question 8
    1 / -0
    In the given figure, if AB ∥ CD and OHF = 30°, then what is the value of x?

    Solution
    In the given figure,
    CEG = 78° (∵ CEG is vertically opposite to 78°)
    CEG = EGH = 78° (∵ Alternate interior angles)
    Now, HFD = 73° (∵ HFD is vertically opposite angle to 73°)
    HFD = FHG = 73° (∵ Alternate interior angles)
    GHO + OHF = FHG
    GHO = 73° - 30° [∵ OHF = 30°(given)]
    GHO = 43°
    Now, in triangle EHG,
    x + 78° + 43° = 180° (Angle sum property of triangle)
    x = 180° - 78° - 43°
    x = 180° - 121°
    x = 59°
  • Question 9
    1 / -0
    In the given figure, ∠ZWX = x, ∠WXZ = y and ∠XOY = z. If y is 40° less than x and ∠ZWO is one-third of x, find the values of x, y and z, respectively.

    Solution


    From the given figure,
    ∠ZWX = x
    And it is given that y = x - 40° ... (i)
    In triangle ZWX,
    x + 46° + y = 180° (Angle sum property of triangle)
    x + 46° + x - 40° = 180° [From (i)]
    2x + 6° = 180°
    x == = 87°
    x = 87°
    y = 87° - 40° = 47°
    Now, ∠ZWO = of x (Given)
    So, ∠ZWO = = 29°
    ∠OWX = 87° - 29° = 58°
    Now, in triangle WOX,
    58° + 47° + ∠WOX = 180° (Angle sum property of triangle)
    ∠WOX = 180° - 105° = 75°
    Now, ∠WOX + ∠YOX = 180° (Linear pair)
    75° + ∠YOX = 180°
    ∠YOX = 180° - 75° = 105°
    z = 105°
  • Question 10
    1 / -0
    In the given figure, if TR ∥ PQ, then what is the value of x?
    Solution


    ∠UTP + ∠PTO + ∠ RTS = 180° ( They lie on a line)

    70° + ∠PTO + 30° = 180°

    ∠ PTO = 180° - 100° = 80

    As TR ∥ PQ, ∠ PTO and ∠ x for co-interior angles and their sum will be 180°.

    ∠ PTO + ∠ x = 180°

    ∠ x = 100°
  • Question 11
    1 / -0
    In the given figure, if OM ∥ LN, then the value of ∠NMO is
    Solution
    In triangle LMN, by angle sum property of triangle,
    3x + 2 + 3(x - 4) + 4(x - 5) = 180°
    3x + 2 + 3x - 12 + 4x - 20 = 180°
    10x - 30° = 180°
    10x = 180°+ 30°
    x == 21°
    Now, ∠LMN + ∠NMO + ∠OMP = 180° (Angles on a straight line)
    3(21 - 4) + ∠NMO + 3 x 21 = 180°
    3 x 17 + ∠NMO + 63° = 180°
    51° + ∠NMO + 63° = 180°
    ∠NMO = 180° - 63° - 51°
    ∠NMO = 66°
  • Question 12
    1 / -0
    In the given figure, if ∠AOC = 90°, find the value of 2∠COD + ∠JOD.

    Solution
    ∵ ∠AOC = 90°
    ∴ ∠AOB + ∠BOC = 90°
    2x - 10 + 3x - 5 = 90°
    5x - 15 = 90°
    5x = 105°
    x = 21°

    Now, ∠COD = 2x + 7 = 2 × 21° + 7 = 42° + 7° = 49°
    2∠COD + ∠JOD = 2 × 49° + × 90° = 98° + 135° = 233°
  • Question 13
    1 / -0
    In the given figure, ST ∥ UV. Find the value of ∠SPO.

    Solution
    ∠UOP + ∠POR + ∠ROV = 180° (Angles on a straight line)

    + x + 20 + x = 180°

    = 180°

    8x + 60° = 180° × 3
    8x = 540° - 60°
    x = = 60°
    ∠SPO = ∠POV = x + 20 + x = 2x + 20 = 2 × 60° + 20 = 140° (Alternate interior angles)
    ∠SPO = 140°
  • Question 14
    1 / -0
    The measure of an angle is 6° less than one-third of its complementary angle. What is the value of the complementary angle?
    Solution
    Let the measure of the angle be x.
    Then, its complementary angle = 90° - x
    According to the given instructions,
    x =(90° - x) - 6°
    3x = 90° - x - 18°
    4x = 72°
    x = 18°
    Complementary angle of x = 90° - 18° = 72°
  • Question 15
    1 / -0
    In the given figure, DE ∥ FG ∥ HI. What is the value of APQ?
    Solution
    It is given that DE ∥ FG ∥ HI.
    Therefore, DE ∥ HI and AB is the transversal.
    HBP = QPB = 50° (Alternate interior angles)
    QPB + APQ = 180° (Linear pair)
    APQ = 180° - 50°
    APQ = 130°
  • Question 16
    1 / -0
    In the given figure, AB || CD. If ∠AEO = 130° and ∠FGD = 105°, then what will be the measure of ∠EOF?

    Solution
    In the given figure,
    ∠FGD = ∠EFG = 105° (Alternate interior angles)
    ∠AEO + ∠FEO = 180° (Linear pair)
    130° + ∠FEO = 180°
    ∠FEO = 180° - 130° = 50°
    In FOE,
    ∠FEO + ∠EFO + ∠EOF = 180° (Angle sum property of triangles)
    50° + ∠EOF + 105° = 180° (∵ ∠EFO = ∠EFG)
    ∠EOF = 180° - 155° = 25°
  • Question 17
    1 / -0
    If the ratio of the measure of an angle to that of its complementary angle is 8 : 10, then what is the measure of the complementary angle?
    Solution
    Let the measure of the angle be y.
    Then, measure of its complementary angle = (90° - y) ... (i)
    Measure of angle : Complementary angle = 8 : 10
    Let the measure of the angle be 8x.
    y = 8x (Using (i))
    So, measure of the complementary angle = 10x
    90° - y = 10x (Using(i))
    90° - 8x = 10x (∵ y = 8x)
    90° = 18x
    x = 5°
    Complementary angle = 10x = 10 × 5 = 50°
  • Question 18
    1 / -0
    If DE || FG , MB||CL and ∠BCG = 117°, then what is the difference between the values of (z + w) and (x + y)?


    Solution
    In triangle BMC,
    44° + x = 117° (An exterior angle is equal to the sum of opposite interior angles.)
    x = 73°
    x = 53° + y
    73° = 53° + y (Alternate interior angles)
    y = 20°
    y + 53° + 44° + z = 180° (Angles on a straight line)
    20° + 53° + 44° + z = 180°
    z + 117° = 180°
    z = 63°
    ∠MBE = 44° + z = 44° + 63° = 107°
    ∠MBE = w (Corresponding angles)
    w = 107°
    (x + y) = 73° + 20° = 93°
    (z + w) = 63° + 107° = 170°
    Difference = 170° - 93° = 77°
  • Question 19
    1 / -0
    In the given figure, if AB || CD, and LM and FH are transversals, find the value of ∠LSB + 4a + 5b.


    Solution
    ∠MSB = 85° (Alternate interior angles)
    ∠MSB + ∠LSB = 180° (Linear pair)
    85° + ∠LSB = 180°
    ∠LSB = 180° - 85°
    ∠LSB = 95°

    a + 85° = 180° (Linear pair)
    a = 95°
    In triangle TRW,
    b + 85° = 165° (Exterior angle property)
    b = 80°
    ∠LSB + 4a + 5b = 95° + 4(95°) + 5(80°) = 875°
  • Question 20
    1 / -0
    In the given figure, what is the measure of ∠WOV?

    Solution
    ∠WOV = (Vertically opposite angles)

    ∠WOP = (Vertically opposite angles)

    x + + + = 180° (Angles on a straight line)

    = 180°

    90x = 180° × 21

    x = = 42°

    x = 42°

    ∠WOV = × 42° = 54°
  • Question 21
    1 / -0
    State 'T' for true and 'F' for false for the following statement.

    (i) The sum of two adjacent angles is 180° if a ray stands on a line.
    (ii) If the sum of two adjacent angles is 180°, then the non-common arms of the angles form a line.
    (iii) Two lines parallel to the same line are not parallel to each other.
    Solution
    (i) When a ray stands on a line, two adjacent angles are formed.
    We know that the angle lying on a straight line is 180°.
    Hence, (i) is true.

    (ii) Angles formed on a straight line are equal to 180 degree. So, the non-common arms of the the angles will form a line.

    Hence, (ii) is true.

    (iii)


    In the above figure, it is given that
    (i)
    (ii)

    So, if and , then .

    Hence, (iii) is false.
  • Question 22
    1 / -0
    Fill in the blanks.

    (a) An exterior angle of a triangle is ________ the sum of opposite interior angles.
    (b) Interior angles on the same side of the transversal are called _______ angles.
    (c) The angle between the bisectors of two right angles on a straight line sharing an arm measures ____.
    Solution
    (i) The exterior angle theorem states that the exterior angle formed when you extend the side of a triangle is equal to the sum of its non-adjacent angles.



    ∠1 + ∠2 = ∠3

    (ii) When a transversal cuts through two parallel lines, the angles on the same side of the transversal are called co-interior angles or consecutive interior angles.

    (iii)


    In the above figure, OQ and OP are the angle bisectors of both the right angles.
    Each of the bisectors is dividing the angle into 45°.
    Therefore, ∠POC = 45°
    And ∠QOC = 45°
    ∠POQ = 45° + 45° = 90°
  • Question 23
    1 / -0
    Which of the following statements is INCORRECT?
    Solution
    Option 1:

    Alternate interior angles are formed when a transversal passes through two lines. The angles that are formed on opposite sides of the transversal and inside the two lines are alternate interior angles.
    In the given figure, SG || RH and L is the transversal.
    a = 58° (Alternate interior angles)
    Hence, the given statement is correct.
    Option 2:
    We know that corresponding angles are equal.

    Let ∠A and ∠B be a pair of corresponding angles.
    If ∠A = 40°, then ∠B = 40°
    Therefore, ∠A + ∠B = 40° + 40° = 80°
    So, sum of corresponding angles may only be equal to 180° when the value of one of the angles is 90°.
    Hence, this statement is incorrect.

    Option 3:

    Let ∠A and ∠B be a linear pair. We know that sum of angles in a linear pair is always 180°.
    Case I: if ∠A = 100°, then ∠B has to be equal to 80°.
    Case II: if ∠A = 60°, then ∠B has to be equal to 120°.
    Case III: If ∠A = 90° then ∠B has to be equal to 90°.

    Option 4:
    We know that vertically opposite angles are those angles which are formed when two lines intersect, and they are equal to each other.


    In the above figure,
    ∠1 = ∠2 and ∠3 = ∠4 (As they are vertically opposite)
  • Question 24
    1 / -0
    In the given figure, ∠QMN = 172° and ∠PMN = 45°. Find the measures of ∠LMO and ∠OMP, respectively.
    Solution
    In the above figure,
    ∠QMN = 172°
    ∠PMN = 45°
    172° = 30° + (3x - 4)° + (x - 3)° + 45°
    172° = 75° + 4x - 7°
    172° = 68° + 4x
    x = 26° ... (I)
    Putting the value of (I) here:
    ∠LMO = (3x - 4)° = 3 × 26 - 4° = 78° - 4° = 74°
    ∠OMP = (x - 3)° = 26° - 3° = 23°
  • Question 25
    1 / -0
    Match the following columns according to the figure given below:

    Column I Column II
    A. Linear pair of angles 1. (∠1 and ∠4), (∠6 and ∠7)
    B. Vertically opposite angles 2. (∠2 and ∠1), (∠13 and ∠15)
    C. Consecutive interior angles 3. (∠3 and ∠11), (∠7 and ∠15)
    D. Corresponding angles 4. (∠4 and ∠10), (∠8 and ∠14)

    In the figure, LM∥NO.
    Solution
    A. A linear pair of angles includes adjacent angles on a straight line. In the given figure, (∠2 and ∠1) and (∠13 and ∠15) are linear pairs of angles. Hence, A corresponds to (2).

    B. Vertically opposite angles share the same vertex and they are on the opposite side of the vertex. In the given figure, (∠1 and ∠4) and (∠6 and ∠7) are vertically opposite angles. Hence, B corresponds to (1).

    C. When two lines are crossed by another line (called transversal), the pairs of angles on one side of the transversal but inside the two lines are called consecutive interior angles. In the given figure (∠4 and ∠10) and (∠8 and ∠14) are consecutive interior angles. Hence, C corresponds to (4).

    D. When two lines are crossed by another line (which is called transversal), the angles at the same location at each intersection are called corresponding angles. In the given figure, (∠3 and ∠11) and (∠7 and ∠15) are corresponding angles. Hence, D corresponds to (3).

    Hence, the correct order is 2, 1, 4, 3.
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