The line with equation x+y=0 coincides with the terminal side of an angle 0 in standard position in quadrant ii.

what is the value of tan0?

tan O = -1.

Step-by-step explanation:

x + y = 0

y = -x

This is a straight line passing through the origin with a slope of -1.

So the angle O in quadrant 2 = 135 degrees.

and it's tangent is -1.

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.