What are the values of ‘a’ and ‘b’?

(the ‘i’ is the imaginary number)

Step-by-step explanation:

Recall that  

so

we have then

Isolating b in the second equation we get

b = 1-a

Replace this value in the first equation

Solving the quadratic equation we get two possible solutions for a  

Replacing these values in b=1-a, we get two possible solutions for b

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.

nlgqer

step-by-step