Question

the
( begin{aligned} A O D &=180-120 B &=60^{circ} end{aligned} )
( operatorname{In} triangle A O C, quad 30^{circ}+45^{circ}+angle A O C=180^{circ} )
( angle A O C=180-75^{circ}=105^{circ} )
( begin{aligned} angle C O D=angle A O D+A O C &=105^{circ}+60^{circ} &=165^{circ} end{aligned} )

# 5. 'O' is the centre of the inscribed circle in a 30° - 60° – 90° triangle ABC right angled at C. If the circle is tangent to AB at D then the angle ZCOD is - (A) 120° (B) 135° (C) 150° (D) 1650 Bb C d re of TE

Solution