Question : Inside a square ABCD, $\triangle$BEC is an equilateral triangle. If CE and BD intersect at O, then $\angle$BOC is equal to:
Option 1: 60°
Option 2: 75°
Option 3: 90°
Option 4: 120°
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Correct Answer: 75°
Solution : Given: Inside a square ABCD, $\triangle$BEC is an equilateral triangle. In $\triangle$BOC, $\angle$BCO = 60° (angle of an equilateral triangle) $\angle$OBC = $\frac{90°}{2}$ = 45° (as the diagonal of a square bisects its angle into two equal parts) So, $\angle$BOC = 180° – (60° + 45°) = 75° Hence, the correct answer is 75°.
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Question : 'O' is the circumcentre of triangle ABC lying inside the triangle, then $\angle$OBC + $\angle$BAC is equal to:
Question : In $\triangle$ABC, if the median AD = $\frac{1}{2}$BC, then $\angle$BAC is equal to:
Question : In $\triangle$ABC, $\angle$B = 35°, $\angle$C = 65° and the bisector of $\angle$BAC meets BC in D. Then $\angle$ADB is:
Question : G is the centroid of $\triangle$ABC. If AG = BC, then measure of $\angle$BGC is:
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