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<AOC =65 and <BOC =25
Hardikkumar112:
Sirplease solve this question; PQRS is a parallelogram. PX bisects angle P and RY bisects angle R. Prove that: (a) angleXPQ = 1/2 of angle P (b) angle SRY = 1/2 of angle R (c) angle SRY= angle SRY= angle RYQ (d) angle XPQ = angle RYQ (e) RY ||XP.
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Heya,
Angle AOB = 90°
Therefore, Angle AOC + Angle COB = 90°
It is given in the question that,
Angle AOC = (2x -5)°
Angle COB = (x - 10)°
Therefore,
(2x - 5)° + (x - 10)° = 90°
=> (3x - 15)° = 90°
=> 3x = 90° + 15°
=> x = 105°/3
=> x = 35°
From this we can find value of Angles;
So,
Angle AOC = (2x -5)°
= (2×35 - 5)°
= (70 - 5)°
= 65°
Angle COB = (x - 10)°
= (35 - 10)°
= 25°
So, Angle AOC = 65° and Angle COB = 25°
Hope this helps....:)
Angle AOB = 90°
Therefore, Angle AOC + Angle COB = 90°
It is given in the question that,
Angle AOC = (2x -5)°
Angle COB = (x - 10)°
Therefore,
(2x - 5)° + (x - 10)° = 90°
=> (3x - 15)° = 90°
=> 3x = 90° + 15°
=> x = 105°/3
=> x = 35°
From this we can find value of Angles;
So,
Angle AOC = (2x -5)°
= (2×35 - 5)°
= (70 - 5)°
= 65°
Angle COB = (x - 10)°
= (35 - 10)°
= 25°
So, Angle AOC = 65° and Angle COB = 25°
Hope this helps....:)
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