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Advanced Triangles Problem QR#161 PS

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by anshumishra » Sun Jan 16, 2011 4:31 pm
tonebeeze wrote:In the figure attached, point O is the center of the circle and OC = AC = AB. What is the value of x (in degrees)?

a. 40
b. 36
c. 34
d. 32
e. 30

OA = B

∠AOC = x = ∠OAC (since, OC = AC)
∠ACB = y = ∠ABC (since, AC = AB)
∠CAB = z

∠ACB = ∠AOC + ∠OAB => y = 2x
Since, OA = OB = radius of the circle ,
so ∠OAB = ∠OBA => x+z = y = 2x
=> x = z

Since, ∠ACB + ∠ABC + ∠CAB = 180
=> 2y + z = 180
=> 4x + x = 180
=> x = 36

B

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Thanks
Anshu

(Every mistake is a lesson learned )
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by nehatandon » Sun Jan 16, 2011 7:23 pm
anshumishra wrote:
tonebeeze wrote:In the figure attached, point O is the center of the circle and OC = AC = AB. What is the value of x (in degrees)?

a. 40
b. 36
c. 34
d. 32
e. 30

OA = B

∠AOC = x = ∠OAC (since, OC = AC)
∠ACB = y = ∠ABC (since, AC = AB)
∠CAB = z

∠ACB = ∠AOC + ∠OAB => y = 2x
Since, OA = OB = radius of the circle ,
so ∠OAB = ∠OBA => x+z = y = 2x
=> x = z

Since, ∠ACB + ∠ABC + ∠CAB = 180
=> 2y + z = 180
=> 4x + x = 180
=> x = 36

B

Image
Nice solution, way better than in OG review book.
Thanks !
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