The tee for the sixth hole on a golf course is 400 yards from the tee. On that hole, Marsha hooked her ball to the left, as sketched below. Find the distance between Marsha’s ball and the hole to the nearest tenth of a yard. Answer any time! :D

  181.8 yd

Step-by-step explanation:

The law of cosines is good for this. It tells you for triangle sides 'a' and 'b' and included angle C, the length of 'c' is given by ...

  c^2 = a^2 +b^2 -2ab·cos(C)

For the given geometry, this is ...

  c^2 = 400^2 +240^2 -2(400)(240)cos(16°) ≈ 33,037.75

  c ≈ √33037.75 ≈ 181.8 . . . yards

Marsha's ball is about 181.8 yards from the hole.

< 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)