1. In the given figure, lines PQ and RS intersect each other at point O; ray OA an
OB bisect |POR and POS respectively. If POA : POB = 2:7, then find SOQ a
|BOQ.
Answers
Answer:
∠SOQ = 40°
∠BOQ = 110°
Step-by-step explanation:
In the given figure, lines PQ and RS intersect each other at point O; ray OA an OB bisect |POR and POS respectively. If POA : POB = 2:7, then find SOQ a
|BOQ.
Let Say ∠POR = x°
∠POS + ∠POR = 180° ( RS Straight line)
=>∠POS + x° = 180°
=> ∠POS = 180° - x°
OA bisects ∠POR
=> ∠POA = (1/2)∠POR = x/2
OB bisects ∠POS
=> ∠POB = (1/2)∠POS = (180 -x)/2 = 90 - x/2
∠POA : ∠POB = 2:7
=> ∠POA / ∠POB = 2/7
=> 7 ∠POA = 2 ∠POB
=> 7 (x/2) = 2 (90 - x/2)
=> 7x/2 = 180 - x
=> 9x/2 = 180
=> x = 40
∠POR = 40°
∠SOQ = ∠POR ( opposite angles)
=>∠SOQ = 40°
∠BOQ =∠BOS + ∠SOQ
=> ∠BOQ = ( 180 - 40)/2 + 40
=> ∠BOQ = 110°
Answer:
POR+∠POS=180
o
.....(1) (Linear pair of angles)
Ir is given that, ray OA and ray OB bisect ∠POR and ∠POS respectively.
Therefore,
∠POA=
2
1
∠POR and ∠POB=
2
1
∠POS
⇒∠POA+∠POB=
2
1
{POR+POS}
=
2
1
×180
o
=90
o
(From equation 1)
Now, if ∠POA+∠POB=2:7
Sum of the ratios =2+7=9
then, we have ∠POA=
9
2
×90
o
=20
o
and ∠POB=
9
7
×90
o
=70
o
∠POR=2×∠POA=2×20
o
=40
o
∠SOQ=∠POR (Vertically opposite angles)
∴∠SOQ=40
o
∠BOQ=∠BOS+∠SOQ
=∠POB+∠SOQ (Since, OB bisect ∠POS)
(∠BOS=∠POB)
=70
o
+40
o
=110
o
∴∠BOQ=110
o