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27 Nov 2019, 08:53
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In the given figure AB is the diameter of the circle, whose center is O. Point C lies inside the circle such that ∠ACB=120 degrees. Also, OD is perpendicular to BC and length of OD is 1.5 cm. Find the length of AC?
A. \(\frac{3\sqrt{3}}{2}\)
B. 3
C. \(2\sqrt{3}\)
D. \(3\sqrt{3}\)
E. 6
A. \(\frac{3\sqrt{3}}{2}\)
B. 3
C. \(2\sqrt{3}\)
D. \(3\sqrt{3}\)
E. 6
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27 Nov 2019, 21:13
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nick1816
Extend BC to meet the circle at E, and join AE.
\(\angle BEA\) is 90 as it is an angle on the diameter AB.
Therefore, AE||OD, as angle AEB and ODB are 90, and AE is double of OD, since BOD and ABE are similar triangles in the ratio 1:2=BO:BA.
Take triangle ACE.
Angle AEB= angle AEC = 90, as it is opposite the diameter AB.
Angle AEC=180-120=60.
Triangle AEC thus is 30-60-90 triangle.
Now AE opposite 60 degree angle is 3, so hypotenuse \(AC=\frac{2*3}{\sqrt{3}}=2\sqrt{3}\), as the sides are in ratio 1:sqrt3:2
C
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27 Nov 2021, 03:05
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chetan2u
how can it be AE||OD?
