Question

question no. 5 and 6​

Answer 1

Question:

5. In fig 6.17, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR.

Prove that ∠ROS = ¹/₂ (∠QOS - ∠POS)

• SOLUTION:

Given:

OR ⊥ PQ  [OR is perpendicular to PQ]

∵ ∠QOR = 90°  

Now,

From the figure, we get that

∠POR and ∠QOR are linear pair(sum = 180°)

⇒ ∠POR + ∠QOR = 180°

⇒ ∠POR + 90° = 180°    

[∵ from given ∠QOR = 90° ]

⇒ ∠POR = 180° - 90°

⇒ ∠POR = 90°

By looking at the figure, we also get that,

∠POR = ∠POS + ∠ROS

⇒ 90° = ∠POS + ∠ROS

⇒ 90° - ∠POS = ∠ROS  -------(i)

→ Also, ∠QOS and ∠POS are linear pairs.

⇒ ∠QOS + ∠POS = 180°

$\implies \frac{1}{2}( \angle QOS+ \angle POS ) = \frac{1}{2} \times 180^{o}$

[Multiplying 1/2 to both the sides]

$\implies \frac{1}{2}( \angle QOS+ \angle POS ) = 90^{o}$  ------(ii)

Now,

Will substitute the eq.(ii) in (i), we get,

$\angle ROS = 90 - \angle POS$

$\angle ROS = [\frac{1}{2}(\angle QOS+ \angle POS)] - \angle POS$

$\angle ROS = \frac{1}{2}(\angle QOS+ \angle POS) - \angle POS$

[Taking LCM 2 in RHS]

$\angle ROS = \frac{\angle QOS+ \angle POS-2 \angle POS)}{2}$

$\angle ROS = \frac{\angle QOS- \angle POS)}{2}$

$\boxed{ \bf \angle ROS = \frac{1}{2} (\angle QOS - \angle POS)} }$

Hence proved.

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Question

6. It is given that ∠XYZ = 64° and XY are produced to point P. Draw the figure from the given information. If ray YQ bisects ∠ZYP, find ∠XYQ and reflex ∠QYP.

• SOLUTION:

[see image 2]

Given -

∠XYZ = 64°

Here,

∠XYZ and ∠ZYP are linear pair (Sum = 180°)

⇒ ∠ XYZ + ∠ZYP = 180°

⇒ 64° + ∠ZYP = 180°

⇒ ∠ZYP = 180° - 64°

⇒ ∠ZYP = 116°

Also,

Ray YQ bisect ∠ZYP (Given)

∠QYZ = ∠QYP

Also,

∠ZYP + 64 = 180

⇒ ∠QYZ + ∠QYP + 64 = 180

⇒ ∠QYZ + ∠QYZ = 180 - 64

⇒ 2∠QYZ = 116

⇒ ∠QYZ = 116 ÷ 2

⇒ ∠QYP = 58°

Now,

∠XYQ = ∠QYZ + ∠XYZ

∠XYQ = 58 + 64

∠XYQ = 112°

So,

Reflex ∠QYP = 360 - ∠QYP

                      = 360 - 58

                      = 302°

Hence,

∠XYZ = 112°

And

Reflex ∠QYP = 302°

Answer 2

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.

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Answer will be (302degrees)

__________________________

It is given that say YQ bisects ∠ZYP

So let, ∠PYQ=∠QYZ=x

∠PYZ=2x

Since QY stands on say PX

∠PYZ+∠ZYX=180

⇒2x+64 =180

[∵∠ZYX=∠XYZ=64 ]

⇒x=(180−64)/2

x=58

∴∠XYQ=∠XYZ+∠ZYQ=64 +58 =122

and reflex

∠QYP=360 −∠QYP=360 −x

∠QYP=360 −58 =302 degree

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