Question

In this figure, LX = 62° angle XYZ = 54 if YO and ZO are the bisectors bisectors of angle XYZ respectively of ∆XYZ and angle YOZ. find angle OZY and angle YOZ ​

Answer 1

$\text{In the given figure it is given that:}$

$\angle{X} = 62^{\circ}\; ;\; \angle{XYZ} = 54^{\circ}$

$\text{We know that sum of angles in a triangle is }180^{\circ}$

$\implies \text{In }\triangle{XYZ}$

$\implies \angle{X} + \angle{XYZ} + \angle{XZY} = 180^{\circ}$

$\implies 62^{\circ} + 54^{\circ} + \angle{XZY} = 180^{\circ}$

$\implies 116^{\circ} + \angle{XZY} = 180^{\circ}$

$\implies \angle{XZY} = 180^{\circ} - 116^{\circ}$

$\implies \angle{XZY} = 64^{\circ}$

$\text{It is also given that }YO \text{ and }ZO \text{ are angle bisectors of }\\\angle{XYZ}\text{ and }\angle{XZY} \text{ respectively}$

$\implies \angle{OYZ} = \dfrac12\times \angle{XYZ}$

$\implies \angle{OYZ} = \dfrac{54}{2}$

$\implies \boxed{\angle{OYZ} = 27^{\circ}}$

$\implies \angle{OZY} = \dfrac12{\angle{XZY}}$

$\implies \angle{OZY} = \dfrac{64}{2}$

$\implies \boxed{\angle{XZY} = 32^{\circ}}$

$\text{We know that sum of angles in a triangle is }180^{\circ}$

$\implies \text{In }\triangle{OYZ}$

$\implies \angle{OYZ} + \angle{OZY} + \angle{YOZ} = 180^{\circ}$

$\implies 27^{\circ} + 32^{\circ} + \angle{YOZ} = 180^{\circ}$

$\implies 59^{\circ} + \angle{YOZ} = 180^{\circ}$

$\implies \angle{YOZ} = 180^{\circ} - 59^{\circ}$

$\implies \boxed{\angle{YOZ = 121^{\circ}}}$