Given mZAOC is a straight angle.
mZBOC = 6x + 29°
mZAOB = 3x + 124°
Find mZBOC:
O
с
A

47°

Step-by-step explanation:

Given that m<BOC = 6x + 29, m<AOB = 3x + 124, and m<AOC is a straight angle, therefore, m<AOC = 180°.

m<BOC +m<AOB = 180°

6x + 29 + 3x + 124 = 180

Solve for x

6x + 3x + 29 + 124 = 180

9x + 153 = 180

Subtract both sides by 153

9x = 180 - 153

9x = 27

Divide both sides by 9

x = 3

m<BOC = 6x + 29

Plug in the value of x

m<BOC = 6(3) + 29

m<BOC = 18 + 29 = 47°

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.