HOW DO YOU SOLVE FOR SEGMENT AC USING TAN, SIN, OR COS WHEN A 90 DEGREE TRIANGLE HAS A 46 DEGREE AND A LEG OF 19.2?

Guest Mar 1, 2012

#1**0 **

i don't really understand your question because which side is the leg on?

anyway, if the leg (19.2) is adjacent to the angle (46), then you'd use cos

so the equation should be: cos(46 degrees)=19.2/x

x=AC

but if the leg is opposite of the given angle, then you'd use sin

sin(46 degrees)=19.2/x

plus, you could use sin and cos reciprocals - sec and csc

just remember SOHCAHTOA

anyway, if the leg (19.2) is adjacent to the angle (46), then you'd use cos

so the equation should be: cos(46 degrees)=19.2/x

x=AC

but if the leg is opposite of the given angle, then you'd use sin

sin(46 degrees)=19.2/x

plus, you could use sin and cos reciprocals - sec and csc

just remember SOHCAHTOA

Guest Mar 1, 2012