Solve for x round to the nearest tenth of degree if u have too

x= ~37.4 degrees

there are lots of ways to solve this triangle

solve with law of sine:

i think i briefly explained it in your last question

im thinking for this one we will solve for measure of angle E then subtract it and 90 from 180

180-90-angle E= x

Pythagorean them for side EG

2.6^2 +3.4^2 = (EG)^2

EG= ~4.28

apply law of sine

call measure angle e =a

3.4 /sin of a= 4.28 /sin 90

3.4*(sin 90) = 4.28* (sin of a)

sin90=1

3.4 = 4.28* (sin of a)

3.4/ 4.28 = (sin of a)

sin of a= ~0.7944

do sin inverse to get value of a

a= 52.5986 degrees

180-90-a= x

180-90-52.5986=x

x=37.4014

x= ~37.4 degrees

hope this helps !!

srry idk

step-by-step explanation:

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)