9类RD Sharma解决方案–第8章直线和角度简介-练习8.2 |套装2

问题11。在图中，ACB是这样的线，∠DCA= 5x，andDCB = 4x。找到x的值。

解决方案：

It is given that ACB is a line in the figure given below.

Thus, ∠ACD and ∠BCD form a linear pair.

Therefore, their sum must be equal toh 180°.

Or we can say that 

∠ACD + ∠BCD = 180°

Also, ∠ACD = 4x and ∠BCD = 5x. 

This further simplifies to:

4x + 5x = 180

9x = 180

x = $\frac{180}{9}$

x = 20°

问题12。在给定的图中，∠POR= 3x和∠QOR= 2x + 10，找到x的值，对于该值，POQ将是一条线。

解决方案：

Here we have POQ as a line

So, ∠POR and ∠QOR form a linear pair.

Therefore, their sum must be equal to 180°

Or

∠POR + ∠QOR = 180°

It is given that ∠POR = (3x)° and ∠QOR = (2x + 10)°. On substituting these values above we get,

3x + (2x + 10) = 180°

3x + 2x + 10 = 180° 

5x + 10 = 180°

5x = 180 – 10

5x = 170

x = $\frac{170}{5}$

x = 34°

问题13。在给定的图中，a大于b的直角的三分之一。找到a和b的值。

解决方案：

It is given that in the figure given below; a is greater than b by one-third of a right angle.

Or we can say that, the difference between a and b is $\frac{1}{3}(90°)$

i.e.

a – b = $\frac{1}{3}(90°)$

a – b = 30° ……..(i)

Also a and b forn a linear pair. Therefore, their sum must be equal to 180°.

We can say that:

a + b = 180° ……….(ii)

On adding (i) and (ii), we get:

2a = 180 + 30

2a = 210

a = $\frac{210}{2}$

a = 105°

On putting a = 105 in (i)

105 – b = 30

-b = 30 – 105

-b = -75

b = 75°

Hence, a = 105° and b = 75°

问题14.如果∠AOC= 4y和∠BOC=(6y + 30)，则y的哪个值会使AOB在给定图中成为一条线

解决方案：

Let us assume, AOB as a straight line.

This makes ∠AOC and ∠BOC  to form a linear pair. Therefore, their sum must be equal to 180°.

We can say that:

∠AOC + ∠BOC = 180°

Also, ∠AOC = 4y  and ∠BOC = 6y + 30. This further simplifies to:

4y + (6y + 30) = 180

10y + 30 = 180

10y = 180 – 30

10y = 150

y = $\frac{150}{10}$

y = 15°

Thus, the value of y = 15° make AOB as a line.

问题15.如果给定的数字， AOF和∠FOG形成线性对。

∠EOB=∠FOC= 90°和∠DOC=∠FOG=∠AOB= 30°

(i)找出∠FOE，∠COB和∠DOE的量度。

(ii)命名所有直角。

(iii)命名三对相邻的互补角。

(iv)命名三对相邻的补充角。

(v)命名三对相邻的角。

解决方案：

The given figure is as follows:

(i) It is given that ∠AOB, ∠FOE, ∠EOB and ∠FOG form a linear pair.

Therefore, their sum must be equal to 180°

i.e.

∠AOB + ∠FOE + ∠EOB + ∠FOG = 180°

It is given that:

∠FOG = 30°

∠AOB = 30°

∠EOB = 90° in equation above, we get:

∠AOB + ∠FOE + ∠EOB + ∠FOG = 180°

30° + ∠FOE + 90° + 30° = 180°

∠FOE + 150° = 180°

∠FOE = 180° – 150° 

∠FOE = 30°

It is given that 

∠FOC = 90°

From the above figure:

∠FOE + ∠DOE + ∠COD = 90°

30° + ∠DOE + 30° = 90°

∠DOE + 60° = 90°

∠DOE = 90° – 60°

∠DOE = 30°

Similarly, we have:

∠EOB = 90°

From the above figure:

∠DOE + ∠DOC + ∠COB = 90°

30° + 30° + ∠COB = 90°

∠COB + 60° = 90°

∠COB = 90° – 60°

∠COB = 30°

(ii) We have:

∠FOG = 30°

∠FOE = 30°

∠EOD = 30°

∠COD = 30°

∠COB = 30°

∠AOB = 30°

From the figure above and the measurements of the calculated angles we get two right angles as ∠DOG and ∠AOD.

Two right angles are already given as ∠FOC and ∠EOB

(iii) We have to find the three pair of adjacent complementary angles.

We know that ∠EOB is a right angle.

Therefore,

∠EOC and ∠COB are complementary angles.

Similarly, ∠AOD is a right angle.

Therefore,

∠AOC and ∠COD are complementary angles.

(iv) We have to find the three pair of adjacent supplementary angles.

Since, ∠AOG is a straight line.

Therefore, following are the three linear pair, which are supplementary:

∠AOB and ∠BOG

∠AOC and ∠COG

∠AOD and ∠DOG

(v) We have to find pair of adjacent angles, which are as follows:

∠AOB and ∠BOC

∠COD and ∠DOE

∠EOF and ∠FOG

问题16.在给定的图中，OP，OQ，OR和OS是四束光线。证明：

∠POQ+∠QOR+∠SOR+∠POS= 360°。

解决方案：

Let us draw TOP as a straight line.

Since, TOP is a line, therefore, ∠POQ, ∠QOR and ∠ROT form a linear pair.

Also, ∠POS and ∠SOT form a linear pair.

Thus, we have:

∠POQ + ∠QOR + ∠ROT = 180° ……(i)

and

∠POS + ∠SOT = 180°  …….(ii)

On adding (i) and (ii), we get;

(∠POQ + ∠QOR + ∠ROT) + (∠POS + ∠SOT) = 180° + 180°

∠POQ + ∠QOR + (∠ROT + ∠SOT) + ∠POS = 360°

∠POQ + ∠QOR + ∠SOR + ∠POS = 360°

Hence proved.

问题17。在给定的图中，ray OS站在POQ线上，Ray OR和ray OT分别是∠POS和angleSOQ的角平分线。如果∠POS= x，则找到∠ROT。

解决方案：

In the figure given below, we have

Ray OR as the bisector of ∠POS

Therefore,

∠POR = ∠ROS

or,

∠POS = 2∠ROS ………..(i)

Similarly, ray OT as the bisector of ∠SOQ

Therefore,

∠TOQ = ∠TOS

or,

∠QOS = 2∠TOS ……….(ii)

Also, Ray OS stand on a line POQ. Therefore, ∠POS and ∠QOS form a linear pair.

Thus,

∠POS + ∠QOS = 180°

From (i) and (ii)

2∠ROS + 2∠TOS = 180°

2(∠ROS + ∠TOS) = 180°

∠ROS + ∠TOS = $\frac{180°}{2}$

∠ROT = 90°

问题18.在给定图中，线PQ和RS在点O处相交。如果∠POR：∠ROQ= 5：7，则找到所有角度。

解决方案：

Let ∠POR and ∠ROQ be 5x and 7x respectively.

Since, Ray OR stand on line POQ. Thus, ∠POR and ∠ROQ form a linear pair.

Therefore, their sum must be equal to 180°.

Or,

∠POR + ∠ROQ = 180°

5x + 7x = 180°

12x = 180°

x = $\frac{180°}{12}$

x = 15° …….(i)

Thus,

∠POR = 5x

= 5(15)

= 75

∠POR = 75°

Thus,

∠ROQ = 7x

= 7(15)

=105

∠ROQ = 105°

It is evident from the figure, that ∠QOS and ∠POR are vertically opposite angles.

And we know that vertically opposite angles are equal.

Therefore,

∠QOS = ∠POR

∠QOS = 75°

Similarlly, ∠POS and ∠ROQ are vertically opposite angles.

And we know that vertically opposite angles are equal.

Therefore,

∠POS = ∠ROQ

∠POS = 105°

问题19.在给定图中，POQ是一条线。射线OR垂直于线PQ。 OS是位于射线OP和OR之间的另一条射线。证明

∠ROS= 1212(∠QOS-POS)。

解决方案：

The given figure shows:

We have POQ as a line. Ray OR is perpendicular to line PQ. Therefore,

∠ROQ = 90°

∠POR = 90°

From the figure above, we get

∠ROS + ∠POS = 90° ………(i)

∠POS and ∠QOS form a linear pair. 

Therefore,

∠QOS + ∠POS = 180°   ……(ii)

From (i) and (ii) equation we get:

∠QOS + ∠POS = 2 × 90 ∠QOS + ∠POS = 2 × 90

∠QOS + ∠POS = 2(∠ROS + ∠POS)

2∠ROS = ∠QOS – ∠POS

∠ROS = $\frac{1}{2}$ (∠QOS – ∠POS)

Hence, proved.