Question

and
Example 2 : In Fig. 6.10, ray OS stands on a line POQ. Ray OR and ray OT are
angle bisectors of Z POS and 2 SOQ, respectively. If Z POS = x, find Z ROT.
the line PO​

Answer 1

Answer:

✯✯ QUESTION ✯✯

In Fig.6.10,Ray OS stands on a line POQ . RayOR and ray OT are angle bisector of angle POS and angle SOQ, respectively . If angle POS=x, find angle ROT.

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✰✰ ANSWER ✰✰

➥Given : -

➥To Find : -

➥Now ,

OR is angle bisector of  

➥So ,

➥Also ,

OT is bisector of  

➥Now ,

90degree is the answer

Step-by-step explanation:

Answer 2

$\huge{\orange{\underline{\purple{\mathscr{Solution}}}}}$

Ray OS stands on the line POQ.

$\:\sf\red{∴}\:$⠀⠀⠀⠀⠀⠀⠀⠀∆POS + ∆SOQ= 180°

$\:\sf\underline\red{But}\:$⠀⠀⠀⠀⠀⠀∆POS = $x$

$\:\sf\red{∴}\:$ ⠀⠀⠀⠀⠀⠀⠀$x$ + ∆SOQ =180°

$\:\sf\underline\red{So\:,}\:$⠀⠀⠀⠀⠀⠀∆SOQ = 180° - $x$

Now ray OR bisects ∆POS , therefore ,

⠀⠀⠀⠀⠀⠀∆ ROS = $\dfrac{1}{2}$ × ∆POS

⠀⠀⠀⠀⠀⠀= $\dfrac{1}{2}$ × $x$ = $\dfrac{x}{2}$

$\:\sf\underline\red{Similarly}\:$⠀⠀⠀⠀∆SOT = $\dfrac{1}{2}$ × ∆SOQ

⠀⠀⠀⠀⠀⠀⠀= $\dfrac{1}{2}$ × (180° -$x$ )

⠀⠀⠀⠀⠀⠀= 90° - $\dfrac{x}{2}$

$\:\sf\underline\red{Now}\:$ ⠀⠀⠀∆ROT = ∆ ROS + ∆ SOT

⠀⠀⠀⠀⠀⠀ = $\dfrac{x}{2}$ +90° - $\dfrac{x}{2}$

⠀⠀⠀⠀⠀⠀= ${\purple{\boxed{\large{\bold{90°}}}}}$