Question #fd0d1

Question

209 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-26T10:35:36

#sin2Q=3/5#

#tanQ=3#

now

tan#=("opposite")/("adjacent")#

#tanQ=3/1#

the hypotenuse of the triangle will then be

#sqrt(3^2+1^2)=10#

#sinQ=2sinQcosQ=2xx3/sqrt10xx1/sqrt10#

#=6/10=3/5#