Question

ABCD is a square. Prove that triangle OAB is an isosceles triangle. Also, find (i) Angle XOD (ii) angle XOC.

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Answer 1

Answer:

It is very easy

First let the angle D and angle O as x° in ∆XOD

Then we know sum of all angles = 180°

So in triangle XOD

78°+x°+x°=180°

2x°=180°-78°

x°= 102°÷2

x°=51°

Mean's angle D and O = 51°

Answer 2

Answer:

diagonals bisect each other in square so OA = OB

as the diagonals bisect the angles angles XDO and XCO= 90°/2 = 45°

according to the angle some property of triangle - sum of angles of a triangle = 180°

i)45° + 78° + angle XOD = 180°

angle XOD = 180° - (45° + 78°) = 57°

ii) Angle DXC = 180° - 78° = 102° ( linear pair)

102° + 45° + angle XOC = 180°

angle XOC = 180° - ( 102° + 45°) = 33°