If two adjacent sides of a cyclic quadrilateral are 2 and 5 and the angle between them is 60 degeree. If the third side is 3,then the remaining fourth side is ? Thanks in advance
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See diagram.
First draw ΔABC with AB= 2, BC=5 and ∠ABC=60°. Then find its circumcenter O and then draw the circle. Then from C draw an arc of radius 3 units to cut the circle at D. Join AD. We have to find AD.
ΔABC: Draw AE ⊥ BC. AE = 2*sin60°= √3
and BE = 2 * cos 60° = 1
So EC = 4
ΔAEC: AC² = AE² + EC² = 19, AE = √19
ΔADC: Draw AF ⊥ CD. Let AD = x. ∠ADC = 120° (180-∠ABC)
ΔADF: ∠ADF = 60° , So AF = x sin60° = √3x/2
FD = x cos60 = x/2
ΔAFC: AF² = AC² - FC²
(√3x/2)² = 19 - (3+x/2)²
3x²/4 = 19 - 9 - x²/4 - 3x
x² + 3x - 10 = 0
(x + 5) (x - 2) = 0
So AD = x = 2
First draw ΔABC with AB= 2, BC=5 and ∠ABC=60°. Then find its circumcenter O and then draw the circle. Then from C draw an arc of radius 3 units to cut the circle at D. Join AD. We have to find AD.
ΔABC: Draw AE ⊥ BC. AE = 2*sin60°= √3
and BE = 2 * cos 60° = 1
So EC = 4
ΔAEC: AC² = AE² + EC² = 19, AE = √19
ΔADC: Draw AF ⊥ CD. Let AD = x. ∠ADC = 120° (180-∠ABC)
ΔADF: ∠ADF = 60° , So AF = x sin60° = √3x/2
FD = x cos60 = x/2
ΔAFC: AF² = AC² - FC²
(√3x/2)² = 19 - (3+x/2)²
3x²/4 = 19 - 9 - x²/4 - 3x
x² + 3x - 10 = 0
(x + 5) (x - 2) = 0
So AD = x = 2
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in a cyclic quadrilateral opposite sides add up to 180 degree.
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