Fundamental Concepts of Geometry — Exercise 14(E)

Two straight lines AB and CD intersect each other at a point O and angle AOC = 50°; find:

(i) angle BOD

(ii) ∠AOD

(iii) ∠BOC

Answer

Given, ∠AOC = 50°.

(i) As, AB and CD intersect each other at point O.

⇒ ∠BOD and ∠AOC are vertically opposite angles.

⇒ ∠BOD = ∠AOC = 50°     [vertically opposite angles are equal]

Hence, ∠BOD = 50°.

(ii) Since CD is a straight line, ∠AOC and ∠AOD form a linear pair.

⇒ ∠AOC + ∠AOD = 180°

⇒ 50° + ∠AOD = 180°

⇒ ∠AOD = 180° − 50° = 130°

Hence, ∠AOD = 130°.

(iii) Since AB is a straight line, ∠AOC and ∠BOC form a linear pair.

⇒ ∠AOC + ∠BOC = 180°

⇒ 50° + ∠BOC = 180°

⇒ ∠BOC = 180° − 50° = 130°

Hence, ∠BOC = 130°.

The adjoining figure shows two straight lines AB and CD intersecting at point P. If ∠BPC = 4x - 5° and ∠APD = 3x + 15°, find:

(i) the value of x.

(ii) ∠APD

(iii) ∠BPD

(iv) ∠BPC.

Answer

(i) As,

AB and CD are intersecting lines at point P.

⇒ ∠BPC and ∠APD are vertically opposite angles.

⇒ ∠BPC = ∠APD     [vertically opposite angles are equal]

⇒ 4x − 5° = 3x + 15°

⇒ 4x − 3x = 15° + 5°

⇒ x = 20°

Hence, x = 20°.

(ii) ∠APD = 3x + 15° = 3 × 20° + 15° = 60° + 15° = 75°

Hence, ∠APD = 75°.

(iii) Since CD is a straight line, ∠APD and ∠BPD form a linear pair.

⇒ ∠APD + ∠BPD = 180°

⇒ 75° + ∠BPD = 180°

⇒ ∠BPD = 180° − 75° = 105°

Hence, ∠BPD = 105°.

(iv) ∠BPC = 4x − 5° = 4 × 20° − 5° = 80° − 5° = 75°

Hence, ∠BPC = 75°.

The figure, given alongside, shows two adjacent angles AOB and AOC whose exterior sides are along the same straight line. Find the value of x.

Answer

From the figure,

∠AOB = 68° and ∠AOC = (3x − 20°).

Since the exterior arms OB and OC of the adjacent angles are in the same straight line, the adjacent angles are supplementary.

⇒ ∠AOB + ∠AOC = 180°

⇒ 68° + (3x − 20°) = 180°

⇒ 3x + 48° = 180°

⇒ 3x = 180° − 48°

⇒ 3x = 132°

⇒ x = $\dfrac{132\degree}{3}$

⇒ x = 44°

Hence, x = 44°.

Each figure, given below, shows a pair of adjacent angles AOB and BOC. Find whether or not the exterior arms OA and OC are in the same straight line.

Answer

The exterior arms OA and OC are in the same straight line only if the sum of the adjacent angles ∠AOB and ∠BOC is 180°.

(i) ∠AOB + ∠BOC = (90° − x) + (90° + x) = 180°

Since the sum is 180°, the exterior arms OA and OC are in the same straight line.

Hence, Yes, OA and OC are in the same straight line.

(ii) ∠AOB + ∠BOC = 97° + 83° = 180°

Since the sum is 180°, the exterior arms OA and OC are in the same straight line.

Hence, Yes, OA and OC are in the same straight line.

(iii) ∠AOB + ∠BOC = 88° + 112° = 200° ≠ 180°

Since the sum is not 180°, the exterior arms OA and OC are not in the same straight line.

Hence, No, OA and OC are not in the same straight line.

A line segment AP stands at point P of a straight line BC such that ∠APB = 5x - 40° and ∠APC = x + 10°; find the values of x and angle APB.

Answer

Since BC is a straight line and AP stands on it at P, ∠APB and ∠APC form a linear pair.

⇒ ∠APB + ∠APC = 180°

⇒ (5x − 40°) + (x + 10°) = 180°

⇒ 6x − 30° = 180°

⇒ 6x = 180° + 30°

⇒ 6x = 210°

⇒ x = $\dfrac{210\degree}{6}$

⇒ x = 35°

Now,

∠APB = 5x − 40° = 5 × 35° − 40° = 175° − 40° = 135°

Hence, x = 35° and ∠APB = 135°.