Question

please fast please help me ​

Answer 1

$\huge\bold\red{Question:-}$

In Fig., ray OS stand on a line POQ Ray OR and ray OT are angle bisectors of ∠POS and ∠SOQ respectively. If ∠POS = x, then find ∠ROT.

$\huge\bold\green{Solution:-}$

Given:-

➵ ∠POR = ∠SOR

➵ ∠QOT = ∠SOT

To Find :-

➵∠ROT = ?

Solution:-

Let ∠POR = ∠SOR = x

∠QOT = ∠SOT = y

Then,

∠POS + ∠QOS = 180° ( angles in the straight line)

⇢ x + x + y + y = 180°

⇢ 2x + 2y = 180°

⇢ 2(x + y) = 180°

⇢ x + y = 180°/2

⇢ x + y = 90°

⇢ ∠ROT = 90°

$\fbox{∠ROT = 90°}$

Answer 2

Step-by-step explanation:

$\huge \red {\boxed{ \bold{ \green{Solution : }}}}$

Have Given :-

Ray OS stands on a line POQ.

and ∠POS = x

$\bold{∠POS + ∠SOQ = 180° ( liner \: pair )}$

$\bold{x + ∠SOQ = 180° \: ( \because \:∠POS = x)}$

$\bold{∠SOQ = 180° - x}$

Ray OR bisects ∠POS,

$\bold{∠ROS = \frac{1}{2} × ∠POS}$

$\bold{ = \frac{1}{2} \times x \: \: ( \because \: ∠POS = x )}$

$\bold{ = \frac{x}{2} }$

OT bisects ∠SOQ

$\bold{∠SOT = \frac{1}{2} \times∠SOQ}$

$\bold{∠SOT = \frac{(180°-x)}{2}}$

$\bold{∠SOT = \frac{180°}{2}} - \frac{x}{2}$

$\bold{∠SOT = 90° - \frac{x}{2} }$

Now,

Finding ∠ROT

$\bold{∠ROT = ∠ROS + ∠SOT}$

$\bold{∠ROT = \frac{x}{2} - 90°- \frac{x}{2} }$

$\bold{∠ROT = \cancel{\frac{x}{2}} - \cancel\frac{x}{2} + 90° }$

$\bold{∠ROT = 90°}$

I hope it is helpful