Question

in fig. 6.40, ∆x=62,∆xyz=54. if YO and ZO are the bisector of∆xyz and ∆xzy respectively of∆xyz find∆OZY and∆YOZ​

Answer 1

Answer:

Answer is 64°

Step-by-step explanation:

X+Y+Z=180°

62°+54°+Z=180°

116+Z=180°

Z=180°-116°

Z=64°

Answer 2

Given:

• ∆x = 62°
• ∆xyz = 54

To Find:

• Find ∆OZY and ∆YOZ

Solution:

We know that the sum of the interior angles of the triangle.

$\circ \: {\boxed{\tt\green{ a + b + c _{(Triangle)} = 180° }}}$

X + XYZ + XZY = 180°

According to Question,

Putting the values to find the Required values,

$\implies$ 62°+ 54° + XZY = 180°

$\implies$ XZY = 64°

∴ ZO is the bisector.

$\implies$ OZY = ½ XZY

$\implies$ OZY = 32°

Hence,

• The value of the OZY is 32°

Similarly, YO is a bisector.

$\implies$ OYZ = ½ XYZ

$\implies$ OYZ = 27° (As XYZ = 54°)

Since, The sum of the interior angles of the triangle.

$\circ \: {\boxed{\tt\green{ OZY + OYZ + YOZ = 180° }}}$

Putting their respective values,

$\colon\implies$ YOZ = 180°- 32°- 27°

$\colon\implies$ YOZ = 121°

Hence,

• The Value of the YOZ is 121°.