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In the figure shown, point O is the center of the semicircle [#permalink]
10 Mar 2008, 10:43

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A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct
100% (00:00) wrong based on 5 sessions

In the figure shown, point O is the center of the semicircle and point points B, C and D lie on the semicircle. If the length of line segment AB = OC, what is the degree measure of angle BAO?

Sum of all the internal angles of a triangle is 180

So For triangle ACO we can write that Angle CAO + Angle COA + Angle ACO = 180..(1)

Angle on a stright line is also 180 degree. So we can also write that Angle COA + Angle COD = 180..(2)

Also Angle ACO is same as Angle BCO.

Since equation 1 and 2 have same RHS, so we can equate LHS. We have Angle CAO + Angle COA + Angle ACO = Angle COA + Angle COD As you can see Angle COA cancels out in both side, so we are left with Angle CAO + Angle ACO = Angle COD => Angle CAO = Angle COD - Angle ACO = 60 - 40 = 20

Angle CAO is same as Angle BAO so it is also 20 degrees.

As both options are required, so I will go with C.

Last edited by abhijit_sen on 10 Mar 2008, 16:42, edited 1 time in total.

bao = boa = x degrees (angles opposite equal sides in a triangle are equal) obc = ocb = a degrees (angles opposite equal sides in a triangle are equal)

from triangle ABO a +a+ (180-x) = 180 (sum of angles in a triangle is 180) => 2a = x

From triangle BCO: x+x+[180-60-a] = 180 => 2x + 120 -a = 180; 2x -a = 60; 4a-a = 60; a = 20.

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