Find obtuse angle between the pair of line.
3x-y+6=0 & y+2x=5​

3x-y+6=0.......(1)

y+2x=5........(2)

[m1]slope of line 1=- coefficient of x/coefficient of y=3

[m2]=]slope of line 2=- coefficient of x/coefficient of y=-2

we have

tan ø=

tan ø=

tan ø=

tan ø=-1

ø=135°

obtuse angle between the pair of line is 135°

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.