Progression...

Number Theory Level pending

In the given figure, measures of A O B , B O C \angle AOB, \angle BOC and C O D \angle COD form an Arithmetic progression . If the smallest angle, A O B = 2 0 \angle AOB = 20^ \circ , find B O C \angle BOC and C O D \angle COD .

6 5 65^ \circ and 9 5 95^ \circ 7 0 70^ \circ and 12 0 120^ \circ 5 0 50^ \circ and 8 0 80^ \circ 6 0 60^ \circ and 10 0 100^ \circ

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1 solution

Rick B
Sep 3, 2014

A O B + B O C + C O D = 18 0 \angle AOB + \angle BOC + \angle COD = 180^\circ and, since it's an AP, B O C = 2 0 + x \angle BOC = 20^\circ + x and C O D = 2 0 + 2 x \angle COD = 20^\circ + 2x

So we have 2 0 + 2 0 + x + 2 0 + 2 x = 6 0 + 3 x = 18 0 3 x = 12 0 x = 4 0 20^\circ + 20^\circ + x + 20^\circ + 2x = 60^\circ + 3x = 180^\circ \implies 3x = 120^\circ \implies x = 40^\circ

So B O C = 2 0 + 4 0 = 6 0 \angle BOC = 20^\circ + 40^\circ = \boxed{60^\circ} and C O D = 2 0 + 8 0 = 10 0 \angle COD = 20^\circ + 80^\circ = \boxed{100^\circ}

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