Which triangle is similar to triangle AEB using the Pieces of Right Triangles Similarity Theorem? Triangle ABC has measure of angle ABC equal to 90 degrees. E is a point on the hypotenuse AC such that the segment BE is perpendicular to AC. D is a point outside the triangle on the right of E so that triangle EDC has the measure of angle EDC equal to 90 degrees. A. Triangle DCE B. Triangle ECB C. Triangle BEC D. Triangle CED

  C. Triangle BEC

Step-by-step explanation:

Point D is not part of triangle ABC, so the "pieces of a right triangle" similarity theorem cannot apply to any figure containing point D. Also, the middle letter of the triangle designator must be the vertex with the right angle.

Only one answer choice remains: Triangle BEC.

a

explanation:

(sin 42)/31. = sinx)/37

sinx = 37 (sin42)/31

= 37× 0.66/31 = 0.787

x= 51.9

other angle opposite to y = 180-51.9-42

= 180- 93.9 =86.1

so

sin42)/31 = sin 86.1)/y

0.021 = 0.997 /y

y= 0.997/0.021 =47.47