Solving the right triangle ‼️ (round to the nearest tenth) can someone me find a, a, and b❓

see explanation

Step-by-step explanation:

The sum of the 3 angles in a triangle = 180°

Subtract the sum of the 2 angles from 180 for A

A = 180° - (50 + 90)° = 180° - 140° = 40°

Use the cosine and sine ratios in the right triangle for a and b

cos50° = =

Multiply both sides by 26

26 × cos50° = a, thus

a ≈ 16.7 ( to the nearest tenth )

sin50° = =

Multiply both sides by 26

26 × sin50° = b, thus

b ≈ 19.9 ( to the nearest tenth )

55? i think maybe

step-by-step explanation:

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.