Lines & Angles - Exercise 17(B)

In the given figure, AOB is a straight line. Find the measure of ∠AOC.

Answer

Since AOB is a straight line, we have:

∠AOC + ∠BOC = 180°

We know from the figure that ∠BOC = 45°

Substituting the value of ∠BOC in above, we get:

⇒ ∠AOC + 45° = 180°

⇒ ∠AOC = 180° - 45°

⇒ ∠AOC = 135°

Thus, the measure of ∠AOC is 135°.

In the given figure, XOY is a straight line. Find

(i) ∠XOP

(ii) ∠YOP

Answer

Since XOY is a straight line, we have:

∠XOP + ∠YOP = 180°

We know from the figure that ∠XOP = (x + 15)° and ∠YOP = (3x + 25)°

Substituting the value of ∠XOP and ∠YOP in above, we get:

⇒ (x + 15)° + (3x + 25)° = 180°

⇒ 4x° + 40° = 180°

⇒ 4x° = 180° - 40°

⇒ 4x° = 140°

⇒ x° = $\dfrac{140^{\circ}}{4}$

⇒ x° = 35°

Now, let's find the measure of each angle by substituting the value of x:

(i) ∠XOP

∠XOP = (x + 15)°

= (35 + 15)°

= 50°

Thus, the measure of ∠XOP is 50°.

(ii) ∠YOP

∠YOP = (3x + 25)°

= (3(35) + 25)°

= (105 + 25)°

= 130°

Thus, the measure of ∠YOP is 130°.

In the given figure, PQR is a straight line. Find the value of x.

Answer

Since PQR is a straight line, we have:

∠PQA + ∠AQB + ∠BQC + ∠CQR = 180°

We know from the figure that,

∠PQA = 60°, ∠AQB = 15°, ∠BQC = x°, ∠CQR = 40°

Substituting the values in above, we get:

⇒ 60° + 15° + x° + 40° = 180°

⇒ 115° + x° = 180°

⇒ x° = 180° - 115°

⇒ x° = 65°

The value of x° is 65°.

In the given figure BC is produced on both sides to points D and E respectively. Find the values of x and y.

Answer

In the figure, DCE is a straight line.

At point B, the angles ∠ABD and ∠ABC form a linear pair.

∴ ∠ABD + ∠ABC = 180°

We know from the figure that ∠ABD = x° and ∠ABC = 115°

Substituting the values in above, we get:

⇒ x + 115° = 180°

⇒ x = 180° - 115°

⇒ x° = 65°

At point C, the angles ∠ACB and ∠ACE form a linear pair.

∴ ∠ACB + ∠ACE = 180°

We know from the figure that ∠ACB = y° and ∠ACE = 132°

Substituting the values in above, we get:

⇒ y° + 132° = 180°

⇒ y° = 180° - 132°

⇒ y° = 48°

The values are x° = 65° and y° = 48°.

In the given figure, AOB is a straight line. If ∠BOC, ∠COD and ∠DOA be in the ratio 2 : 3 : 4, find the measure of each of these angles.

Answer

Given:

∠BOC = (2x)°

∠COD = (3x)°

∠DOA = (4x)°

Since, AOB is a straight line,

∴ ∠BOC + ∠COD + ∠DOA = 180°

Substituting the values in above, we get:

2x° + 3x° + 4x° = 180°

⇒ 9x° = 180°

⇒ x° = $\dfrac{180^{\circ}}{9}$

⇒ x° = 20°

Let's find the measure of each angle by substituting the value of x:

∠BOC = (2x)° = (2 x 20)° = 40°

∠COD = (3x)° = (3 x 20)° = 60°

∠DOA = (4x)° = (4 x 20)° = 80°

∠BOC = 40°, ∠COD = 60° and ∠DOA = 80°.

In the given figure, find the value of x.

Answer

From the figure,

∠COB = 100°, ∠BOA = x°, ∠AOD = 150°, ∠DOC = 30°

At point O, all angles around a point add up to 360°

∴ ∠COB + ∠BOA + ∠AOD + ∠DOC = 360°

⇒ 100° + x° + 150° + 30° = 360°

⇒ 280° + x° = 360°

⇒ x° = 360° - 280°

⇒ x° = 80°

The value of x° is 80°.

In the given figure, find the measure of each of the angles ∠AOB, ∠BOC, ∠COD and ∠DOA.

Answer

From the figure,

∠AOB = x°, ∠BOC = 2x°, ∠COD = 3x°, ∠DOA = 4x°

At point O, all angles around a point add up to 360°

∴ ∠AOB + ∠BOC + ∠COD + ∠DOA = 360°

⇒ x° + 2x° + 3x° + 4x° = 360°

⇒ 10x° = 360°

⇒ x° = $\dfrac{360^{\circ}}{10}$

⇒ x° = 36°

Let's find the measure of each angle by substituting the value of x:

∠AOB = x° = 36°

∠BOC = 2x° = (2 x 36)° = 72°

∠COD = 3x° = (3 x 36)° = 108°

∠DOA = 4x° = (4 x 36)° = 144°

∠AOB = 36°, ∠BOC = 72°, ∠COD = 108° and ∠DOA = 144°.

In the given figure, find the measure of each of the angles ∠DOE and ∠EOA.

Answer

From the figure,

∠AOB = 90°, ∠BOC = 112°, ∠COD = 86°, ∠DOE = x°, ∠EOA = 3x°

At point O, all angles around a point add up to 360°

∴ ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 360°

⇒ 90° + 112° + 86° + x° + 3x° = 360°

⇒ 288° + 4x° = 360°

⇒ 4x° = 360° - 288°

⇒ 4x° = 72°

⇒ x° = $\dfrac{72^{\circ}}{4}$

⇒ x° = 18°

Let's find the measure of each angle by substituting the value of x:

∠DOE = x° = 18°

∠EOA = 3x° = (3 x 18)° = 54°

∠DOE = 18°, ∠EOA = 54°.

In the given figure, find the value of x. What is the measure of ∠COD?

Answer

From the figure,

∠AOB = (x - 5)°, ∠BOC = (2x - 5)°, ∠COD = (3x + 20)°, ∠DOA = 4x°

At point O, all angles around a point add up to 360°

∴ ∠AOB + ∠BOC + ∠COD + ∠DOA = 360°

⇒ (x - 5)° + (2x - 5)° + (3x + 20)° + 4x° = 360°

⇒ x° + 2x° + 3x° + 4x° - 5° - 5° + 20° = 360° $\quad$[Grouping like terms]

⇒ 10x° + 10° = 360°

⇒ 10x° = 360° - 10°

⇒ 10x° = 350°

⇒ x° = $\dfrac{350^{\circ}}{10}$

⇒ x° = 35°

Let's find ∠COD by substituting the value of x:

∠COD = (3x + 20)° = (3(35) + 20)° = (105 + 20)° = 125°

The value of x° is 35°, and the measure of ∠COD is 125°.

In the given figure, two straight lines AB and CD intersect at a point O. If ∠BOD = 40°, find the measure of each of the angles, ∠BOC, ∠AOC and ∠AOD.

Answer

Given:

∠BOD = 40°

When two lines intersect, the angles opposite to each other are equal.

∠AOC = ∠BOD $\quad$ [Vertically opposite angles]

∴ ∠AOC = 40°

Since AOB is a straight line, ∠BOC and ∠BOD form a linear pair and must sum to 180°

∠BOC + ∠BOD = 180°

Substituting the value of ∠BOD in above, we get:

⇒ ∠BOC + 40° = 180°

⇒ ∠BOC = 180° - 40°

⇒ ∠BOC = 140°

∠AOD = ∠BOC $\quad$ [Vertically opposite angles]

∴ ∠AOD = 140°

∠AOC = 40°, ∠BOC = 140° and ∠AOD = 140°.

In the given figure, two lines AB and CD intersect at a point O. If ∠AOC + ∠BOD = 70°, find the measure of ∠AOD.

Answer

In the given figure, two straight lines AB and CD intersect at point O.

When two lines intersect, the angles opposite to each other are equal:

∴ ∠AOC = ∠BOD $\quad$ [Vertically opposite angles]

The question states that:

∠AOC + ∠BOD = 70°

Since they are equal, we can replace ∠BOD with ∠AOC in the equation:

⇒ ∠AOC + ∠AOC = 70°

⇒ 2∠AOC = 70°

⇒ ∠AOC = $\dfrac{70^{\circ}}{2}$

⇒ ∠AOC = 35°

So, ∠AOC = 35° and ∠BOD = 35°.

Since CD is a straight line, the angles ∠AOC and ∠AOD form a linear pair and must sum to 180°.

∴ ∠AOC + ∠AOD = 180°

⇒ 35° + ∠AOD = 180°

⇒ ∠AOD = 180° - 35°

⇒ ∠AOD = 145°

The measure of ∠AOD is 145°.

In the given figure, the lines AB, CD and EF intersect at a point O. If ∠BOD = x°, ∠AOE = 2x° and ∠COF = 90°, find ∠AOE and ∠AOC.

Answer

Given:

∠BOD = x°, ∠AOE = 2x°, ∠COF = 90°

In the figure, three straight lines (AB, CD, and EF) intersect at point O.

Therefore, vertically opposite angles are:

∠BOD = ∠AOC. Therefore, ∠AOC = x°

∠AOE = ∠BOF. Therefore, ∠BOF = 2x°

∠COF = ∠DOE. Therefore, ∠DOE = 90°

Since AB is a straight line, the angles ∠AOE, ∠EOD and ∠DOB lie on a straight line and their sum is 180°.

∴ ∠BOD + ∠DOE + ∠AOE = 180°

Substituting in the above equation, we get:

x°+ 90° + 2x = 180°

⇒ 3x° + 90° = 180°

⇒ 3x° = 180° - 90°

⇒ 3x° = 90°

⇒ x° = $\dfrac{90^{\circ}}{3}$

⇒ x° = 30°

Let's find ∠AOE and ∠AOC by substituting the value of x:

∠AOE = 2x° = (2 x 30)° = 60°

∠AOC = x° = 30°

∠AOE = 60° and ∠AOC = 30°.