Advertisements
Advertisements
Question
In the given figure, BAC is a straight line.
Find:
(i) x
(ii) ∠AOB
(iii) ∠BOC

Advertisements
Solution
∵ ∠AOB and ∠COB are linear pairs
∴ ∠AOB + ∠COB = 180°
⇒ x + 25° + 3x + 15° = 180°
⇒ 4x + 40° = 180°
⇒ 4x = 180°− 40° = 140°
(i) ⇒ x =`(140°)/4=35°`
Hence, x = 35°
(ii) ∠AOB = x + 25° = 35° + 25° = 60°
(iii) ∠BOC = 3x + 15° = 3 × 35° + 15°
= 105° + 15° = 120°
APPEARS IN
RELATED QUESTIONS
If an angle is 30° more than one half of its complement, find the measure of the angle.
In the given below fig, rays OA, OB, OC, OP and 0E have the common end point O. Show
that ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 360°.

Fill in the blank so as to make the following statement true:
If one angle of a linear pair is acute, then its other angle will be _____
Define supplementary angles.
Define adjacent angles.
In the given figure, if CP || DQ, then the measure of x is

How many rays can be drawn through a fixed point O?
Write the complement of `1/2` of a right angle
Write the complement of 21° 17'
Write the supplement of (110 – x – 2y)°
