有兩條math

1.
Let x=CE, y=AE

兩個三角形, 兩條式
x/y = tan 45

(x+10)/y = tan 70

then solve x and y from 兩條式.
answer is: x=10/(tan70 - 1)

2.
同上, 兩個三角形, 兩條式
let x = MQ, y = AQ

y/x = tan 45

y/(x + 1000) = tan 32

then solve x and y from 兩條式.

12.
tan45=ce/ae=1,
tan70=cd/bd,
ae=bd,
cd=ce+10,
then,
cd-10=ae=cd/tan70,
so,
cd(1-1/tan70)=10
cd=15.723m

13.
tan45=qa/mq=1,
tan32=qa/(mq+1000),
then
tan32=mq/(mq+1000)
mq*tan32+1000tan32=mq
mq(1-tan32)=1000tan32
mq=1666m