ABCD is a square. A line segment DX meets the side BC at X and the diagonal AC at O such that angle COD = 105*. Find angle OXC.
Question
ABCD is a square. A line segment DX meets the side BC at X and the diagonal AC at O such that angle COD = 105°.
Find angle OXC.
Solution
Sure, here are the steps to find the angle OXC:
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Since ABCD is a square, the diagonal AC bisects angle BAD, making angle BAC = 45 degrees.
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Since angle COD = 105 degrees, and angle BAC = 45 degrees, angle BAO = 105 - 45 = 60 degrees.
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Triangle BAO is an isosceles triangle (since AB = AO), so angle ABO = angle BAO = 60 degrees.
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Therefore, angle ABC = 180 - 60 = 120 degrees.
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Since DX is a straight line, angle ABX = 180 - 120 = 60 degrees.
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Since triangle ABX is an isosceles triangle (since AB = AX), angle AXB = angle ABX = 60 degrees.
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Therefore, angle OXC = 180 - 60 = 120 degrees.
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