When she was younger, Mary had to look up at a 68 degree angle to see into her father's eyes whenever she was standing 15 inches away. How high above the flat ground were her father's eyes if Mary's eyes were 32 inches above the ground?

69 INCHES

Step-by-step explanation:

Given that :

Height of father's eye above the flat ground :

Kindly check attached picture for diagram:

To obtain the value of h ; we can use Pythagoras rule ;

Tan θ = opposite / Adjacent

Tan 48 = h / 15

15 tan 48 = h

37.126302 = h

h = 37.

Hence, height of father's above the flat ground :

h + 32 = 37 + 32= 69 inches

answer: .75 km

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)