In right teanglie ABC shown below what is the degree measure of \_ C!!

please help me

20 use https://www.calculator.net/triangle-calculator.html

Step-by-step explanation:

Acute Isosceles Triangle

Side a = 1

Side b = 1

Side c = 0.51764

Angle ∠A = 75° = 1.309 rad = 5/12π

Angle ∠B = 75° = 1.309 rad = 5/12π

Angle ∠C = 30° = 0.5236 rad = π/6

C=30°B=75°A=75°b=1a=1c=0.518

Area = 0.25

Perimeter p = 2.51764

Semiperimeter s = 1.25882

Height ha = 0.5  

Height hb = 0.5

Height hc = 0.96593

Median ma = 0.61966  

Median mb = 0.61966

Median mc = 0.96593

Inradius r = 0.1986  

Circumradius R = 0.51764  

Vertex coordinates: A[0, 0] B[0.51764, 0] C[0.25882, 0.96593]

Centroid: [0.25882, 0.32198]

Inscribed Circle Center: [0.25882, 0.1986]

Circumscribed Circle Center: [0.25882, 0.44829]

+ Show Calculation Steps

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

answer:

f(5/2) = 10/3

i hope this you!

< 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)