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.

Answer 1

Answer:

Step-by-step explanation:

Answer 2

✯✯ 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.

━━━━━━━━━━━━━━━━━━━━

✰✰ ANSWER ✰✰

➥Given : -

• $\implies\tt{\angle{POS}=x}$

➥To Find : -

• $\implies\tt{\angle{ROT}=?}$

➥Now ,

• OR is angle bisector of $\angle{POS}$

➥So ,

$\implies\tt{\angle{POR}=\dfrac{x}{2}\:and\:\angle{ROS}=\dfrac{x}{2}}$

➥Also ,

$\implies\tt{\angle{POS}+\angle{SOQ}=180\degree}$

$\implies\tt{x+\angle{SOQ}=180\degree}$

$\implies\tt{\angle{SOQ}=180\degree-x}$

• OT is bisector of $\angle{SOQ}$

$\implies\tt{\angle{SOT}=\dfrac{180-x}{2}\:and\:\angle{TOQ}=\dfrac{180-x}{2}}$

➥Now ,

$\implies\tt{\angle{ROT}=\angle{ROS}+\angle{SOT}}$

$\implies\tt{\angle{ROT}=\dfrac{x}{2}+\dfrac{180-x}{2}}$

$\implies\tt{\angle{ROT}=\dfrac{x+180-x}{2}}$

$\implies\tt{\angle{ROT}=\cancel\dfrac{180}{2}}$

$\implies\tt\bold{\angle{ROT}=90\degree}$