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# In Fig. 6.40, ∠X = 62°, ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of △XYZ, find ∠OZY and ∠YOZ.

**Solution:**

Given: ∠X = 62°, ∠XYZ = 54°, and YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively.

To find: ∠OZY and ∠YOZ

According to the angle sum property of a triangle, sum of the interior angles of a triangle is 180°.

Consider △XYZ

∠X + ∠XYZ + ∠Z = 180° [Angle sum property of a triangle]

62° + 54° + ∠Z = 180°

∠Z = 180° - 116°

∠Z = 64°

Now, OZ is the angle bisector of ∠XZY

Thus, ∠OZY = (1/2) of ∠XZY = 1/2 × 64° = 32° ..........(i)

Similarly, OY is the angle bisector of ∠XYZ

Thus, ∠OYZ = (1/2) of ∠XYZ = 1/2 × 54° = 27°......... (ii)

Now, in △OYZ

∠OYZ + ∠OZY + ∠YOZ = 180° [Angle sum property of a triangle]

27° + 32° + ∠YOZ = 180° [from (i) and (ii)]

∠YOZ = 180° - 59°

∠YOZ = 121°

Hence, ∠OZY = 32° and ∠YOZ = 121°

**☛ Check: **NCERT Solutions Class 9 Maths Chapter 6

**Video Solution:**

## In Fig. 6.40, ∠X = 62°, ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of △XYZ, find ∠OZY and ∠YOZ

NCERT Maths Solutions Class 9 Chapter 6 Exercise 6.3 Question 2

**Summary:**

In Fig. 6.40, if ∠X = 62°, ∠XYZ = 54°, and YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ΔXYZ, then ∠OZY = 32°, and ∠YOZ = 121°.

**☛ Related Questions:**

- In Fig. 6.41, if AB || DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.
- In Fig. 6.42, if lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.
- In Fig. 6.43, if PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65° then find the values of x and y.
- In Fig. 6.44, the side QR of ∠PQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR = 1/2 ∠QPR.

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