Q1. In a rt. ΔABC, ∠A = 90° and AB = AC. What are the values of ∠B and ∠C?
Sol. ∵ AB = AC
⇒ ∠B = ∠C
Also, ∠A = 90°
⇒ ∠B + ∠C = 90°
⇒ ∠B = ∠C = (90^{o}/2) = 45°
Q2. In the figure, what is the value of x?
Sol. ∵ ℓ  m and p is a transversal
∴ ∠1 + 70° = 180° [cointerior angles]
⇒ ∠1 = 180°  70° = 110°
Now, 2x = 110° [vertically opposite angles]
⇒ x =(110°/2) = 55°
Q3. If two interior angles on the same side of a transversal intersecting two parallel lines are in the ratio 2 : 3 then, what is the smaller angle?
Sol. ℓ  m and p is the transversal
∴ ‘a’ and ‘b’ are interior angles on the same side of the transversal p.
Let a = 2x and b = 3x
∴ a + b = 180°
⇒ 2x + 3x = 180°
⇒ 5x = 180°
⇒ x = (180/5)= 36°
∴ smaller angle = 2x = 2 x 36 = 72°
Q4. In the following figure AB  CD. Find the measure of ∠BOC.
Sol. Extending AB to intersect OC, we get the following figure.
ABF is a straight line
∴ ∠OBF = 180°  165° = 15°
AB  CD ⇒ EF  CD
∴ ∠1 + 75° = 180°
⇒ ∠1 = 180°  75° = 105°
⇒ ∠2 = 105°
Now, in Δ, ∠2 + 15° + ∠BOC = 180°
⇒ 105° + 15° + ∠BOC = 180°
⇒ ∠BOC = 180°  105°  15°
= 60°
Q1. In the given figure, AB, CD, and EF are three lines concurrent at O. Find the value of y.
Sol:
∵ ∠AOE and ∠BOF and vertically opposite angles.
∴ ∠AOE = ∠BOF = 5y ..... (1)
Now, CD is a straight line,
⇒ ∠COE + ∠EOA + ∠AOD = 180°
⇒ 2y + 5y + 2y = 180° [From (1)]
⇒ 9y = 180°⇒ y = (180°/2)= 20°
Thus, the required value of y is 20°.
Q2. In the adjoining figure, AB  CD and PQ is transversal. Find x.
Sol:
∵ AB  CD and PQ is a transversal.
∴ ∠ BOQ = ∠ CQP [∵ Alternate angles are equal]
⇒ x = 110° [∵ ∠ CQP = 110°]
Q3. Find the measure of an angle that is 26° more than its complement.
Sol:
Let the measure of the required angle be x.
∴ Measure of the complement of x° = (90°  x)
⇒ x°  (90°  x) = 26°
⇒ x  90° + x = 26°
⇒ 2x = 26° + 90° = 116°
⇒ x = (116°/2) = 58°
Thus, the required measure = 58°.
Q4. Two supplementary angles are in the ratio 3:2. Find the angles.
Sol:
Let the measure of the two angles be 3x and 2x.
∵ They are supplementary angles.
∴ 3x + 2x = 180°
⇒ 5x = 180° ⇒ x = (180°/5) = 36°
∴ 3x = 3 x 36° = 108° and 2x = 2 x 36° = 72°
Thus, the required angles are 108° and 72°.
Q5. In the adjoining figure, AOB is a straight line.
Sol:
∵ AOB is a straight line.
∴ ∠ AOC + ∠ BOC = 180°
⇒ (3x + 10°) + (2x  30°) = 180° [Linear pair]
⇒ 3x + 2x + 10°  30° = 180°
⇒ 5x  20° = 180°
⇒ 5x = 180° + 20° = 200° ⇒ x = (200°/5) = 40°
Thus, the required value of x is 40°.
Q6. In the adjoining figure, find ∠ AOC and ∠ BOD.
Sol:
∵ AOB is a straight line.
∴ ∠AOC + ∠COD + ∠DOB =180°
⇒ x + 70° + (2x  25°) = 180°
⇒ x + 2x = 180° + 25°  70°
⇒ 3x = 205°  70° = 135° ⇒ x = (135°/3) = 45°
∴ ∠ AOC = 45°
⇒ ∠ BOD = 2x  25° = 2 (45°)  25° = 90°  25° = 65°
Q7. In the adjoining figure, AB  CD. Find the value of x.
Sol:
Let us draw EF  AB and passing through point O.
∴ EF  CD and CO is a transversal.
⇒ ∠ 1 = 25° [Alternate angles]
Similarly, ∠ 2 = 35°
Adding, ∠ 1 + ∠ 2 = 25° + 35° ⇒ x = 60°
Thus, the required value of x is 60°.
Q1. In the adjoining figure AB  CD  EG, find the value of x.
Sol:
Through E, let us draw FEG  AB  CD.
Now, since FE  AB and BE is a transversal.
∴ ∠ ABE + ∠ BEF = 180°
[Interior opposite angles]
⇒ 127° + ∠ BEF = 180°
⇒ ∠ BEF = 180° ∠ 127° = 53°
Again, EG  CD and CE is a transversal.
∴ ∠ DCE + ∠ CEG = 180° [Interior opposite angles]
⇒ 108° + ∠ CEG = 180° ⇒ ∠ CEG = 180°  108° = 72°
Since FEG is a straight line, then
⇒ ∠BEF + ∠BEC + ∠CEG = 180°
[Sum of angles at a point on the same side of a line = 180°]
⇒ 53° + x + 72°
= 180°
⇒ x = 180°  53°  72°
= 55°
Thus, the required measure of x = 55°.
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