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Two complementary angles are such that two times the measure of one is equal to three times the measure of the other. The measure of the larger angle is:
Let the measure of angle be x.
∴ Its complement = (90°  x)
∴ 2x = 3(90°  x)
⇒ 5x = 3 × 90°
⇒ x = 54°
and (90°  x ) = 36°
∴ Measure of larger angle = 54°
In the figure, OA and OB are opposite rays ∠AOC + ∠BOD = 63°. The measure of angle ∠COD is;
∵ OA and OB are opposite rays.
∴ ∠AOB is a straight angle.
⇒ ∠AOB = 180°
⇒ (∠AOC + ∠BOD) + ∠COD = 180°
⇒ 63° + ∠COD = 180°
⇒ ∠COD = 117°
Determine the value of x from the given figure;
∠AOB = 180°
⇒ ∠AOP + ∠POB = 180°
⇒ ∠AOP + ∠POQ + ∠BOQ = 180°
⇒ 90° + 3x + x = 180°
⇒ 4x = 90°
⇒ x = 90°/4 = 22.5°
∵ Sum of angles around a point = 360°
∠AOB + ∠BOC = 360°
⇒ 90° + a + b = 360°
⇒ a + b = 270° ...(i) and
b = a + 20° ...(ii)
Using (ii) in (i)
a + (a + 20°) = 270°
⇒ 2a = 250°
⇒ a = 125°
∠m = ∠x [Vertically opposite ∠s]
∵ ∠AOB = 180°
⇒ ∠BOF + ∠COF + ∠AOC = 180°
⇒ ∠BOF + ∠DOE +∠AOC = 180°
⇒ x° + 2x° + 3x° = 180°
⇒ 6x° = 180°
⇒ x = 30°
∴ m = 30°
Here ∠COQ = ∠POD [vertically opposite ∠s]
∵ ∠AOB = 180° (AOB is a straight line)
⇒ ∠POA + ∠POD + ∠BOD = 180°
⇒ 2x° + 3x° + 20° + 3x° = 180°
⇒ 8x = 160°
⇒ x = 20°
x = 3y, z = 21/6 y = 7/2y.
∴ x + y + z = 180°
⇒ 3y + y + 7/2 y = 180°
⇒ 4y + 7/2 y = 180°
⇒ 15y = 180° × 2
⇒ y = 24°
∠AOD = ∠BOC [vertically opposite ∠s]
⇒ y = 60° + x ...(i) and,
∵ ∠DOC = 180°
⇒ 60° + x + 100° = 180°
⇒ x = 20°
It is clear from the figure that,
105° + 5x = 180°
⇒ 5x = 75°
⇒ x = 15°
AB  CD, and ∠RQB = 115°, and ∠PRQ = 30°. The measure of ∠APC is:
Here ∠RQB + ∠RQP =180° (∵ AB is a straight line)
⇒ ∠RQP = 180°  115° = 65°
Now ∠PRQ = 30°
∵ ∠PRQ, ∠RQP and ∠APQ are the ∠S of D
∴ ∠PRQ + ∠RQP + ∠APQ = 180°
⇒ ∠APQ = 180° 65° 30° = 85°
∠APC = ∠APQ [Vertically opposite ∠S]
∴ ∠APC = 85°
If x : y = 2 : 3, then the value of y is equal to:
∵ X : Y = 2 : 3
∴ Let the angles x and y be 2k and 3k respectively
∴ 2k + 3k = 180°
[Sum of ∠S in the interior of transversal]
⇒ 5k = 180° ⇒ k = 36°
∴ y = 3k = 3 × 36° = 108°
AB  CD  EF and GH  KL. The measure of ∠HKL is
Extending GH to M, we have,
∠CHG = ∠APH = 60° [Corresponding ∠S]
∠APH = ∠MPD = 60° [Vertically opposite ∠S]
∠APH = ∠MPD, then,
∠MPD + ∠LKP = 180° [Sum of interior ∠S]
⇒ ∠LKP = 180°  60° = 120°, also
∠KHD = ∠PKH = 25° (Alternate ∠S)
∴ ∠HKL = ∠LKP + ∠PKH
= 120° + 25° = 145°
Construct a line l  AB,
∠a = 25° [ Alternate ∠S]
∠c = 105°
∴ ∠b = 105° [Vertically opposite ∠S]
∴ x = a + b = 25° + 105° = 130°
∠DD_{1}S = ∠B_{1}D_{1}C_{1} = 2x [Vertically opposite ∠S]
∠B_{1}D_{1}C_{1} + ∠A_{1}B_{1}D_{1} = 180° [Interior ∠S]
⇒ 2x + 2y = 180°
⇒ x + y = 90° ...(i)
Also,
∠AA_{1}C_{1 }+ ∠A_{1}B_{1}D_{1} [Corresponding ∠S]
⇒ 30° = 2y
⇒ y = 15° ...(ii)
Using (ii) in (i), we get
x = 90°  y = 90°  15° = 75°
From the figure,
(2y + y + y) + 35° = 180° (Interior ∠S)
⇒ 5y = 180°  35°
⇒ y = 36°  7
= 29°
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