5) -5+ csc 0 = V3 - 5 + 2csc 0 %3D 2n A) 3 3 B) { 6 6 4π 5π C) { 3 3 2л 5л D) { 3 6.

Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
Chapter5: Similar Triangles
Section5.3: Proving Triangles Similar
Problem 43E
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3) -5+-
· csc 0 = V3 –- 5 + 2csc 0
%3D
n 2n
A)
3 3
元 S元
B) {-
6 6
4n 5T
C) {
3 3
2π5π
D)
6.
5) -6-2cos 0 =-4 + 2cos e
%3D
T ST
A) {
4 4
B) (플
3 3
2n 4n
3'
Transcribed Image Text:3) -5+- · csc 0 = V3 –- 5 + 2csc 0 %3D n 2n A) 3 3 元 S元 B) {- 6 6 4n 5T C) { 3 3 2π5π D) 6. 5) -6-2cos 0 =-4 + 2cos e %3D T ST A) { 4 4 B) (플 3 3 2n 4n 3'
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