What is the vaule of angle(ABC) angle (CBE) in the following figure
In the below figure AB = 9 cms BC = 3 cms CE = 14 cms EF = 4 cm. If the line CE touches the circle at D. Find the lengths of (a) CD (b) ED (c) FG
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.in first one angle at centre is twice at perimetre.so this is logic.in second one use the simple formula. answerof first one is 50 and 130.
Angle (APC) = Angle (ADC) = Half of angle AOC = 50 Angle made at center is twice the angle
Now angle (ABC) + angle (APC) = 180 (sum of opposite sides of a circular quadrilateral) => Angle (ABC) = 130
And angle (CBE) + angle (ABC) = 180 So angle (CBE) = 50
CB x CA = CD2
=> CD2 = 3 x 12 = 36
=> CD = 6
As CE = 14
=>DE = 8
Again DE2 = EF x EG
=> EG = 64/4 = 16
=> GF = 16 - 4 = 12
Q1
Take major arc APDC. Angle subtended by this arc at centre = 360-100 = 260.
therefore angle subtended by this arc at point B i.e angle ABC = 130.
Obviously angle CBE is 50.
the ans is 130 and 50 deg.
the answers are 6 , 8 , 12 respectively
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.in first one angle at centre is twice at perimetre.so this is logic.in second one use the simple formula. answerof first one is 50 and 130.
n/a
Angle (APC) = Angle (ADC) = Half of angle AOC = 50
Angle made at center is twice the angle
Now angle (ABC) + angle (APC) = 180 (sum of opposite sides of a circular quadrilateral)
=> Angle (ABC) = 130
And angle (CBE) + angle (ABC) = 180
So angle (CBE) = 50
CB x CA = CD2
=> CD2 = 3 x 12 = 36
=> CD = 6
As CE = 14
=>DE = 8
Again DE2 = EF x EG
=> EG = 64/4 = 16
=> GF = 16 - 4 = 12
Q1
Take major arc APDC. Angle subtended by this arc at centre = 360-100 = 260.
therefore angle subtended by this arc at point B i.e angle ABC = 130.
Obviously angle CBE is 50.
n/a
the ans is 130 and 50 deg.
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the answers are 6 , 8 , 12 respectively
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