Question

Help!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 1

with what

Answer 2

im new to trig and i have a placement test tomorrow for pre algebra algebra and trig and need some help

Answer 3

Im not the best peson to help sorry ask @phi

Answer 4

You might want to look at
http://www.khanacademy.org/math/trigonometry/basic-trigonometry/basic_trig_ratios/v/basic-trigonometry
and then this
http://www.khanacademy.org/math/trigonometry/basic-trigonometry/basic_trig_ratios/v/example--trig-to-solve-the-sides-and-angles-of-a-right-triangle

Answer 5

In your problem, side BC is opposite the angle 60º
they tell you the hypotenuse (opposite the 90º angle) is 8

using SOH CAH TOA O and H means use SOH
short for
sin 60 = opposite/hypotenuse = BC/8
\[ \sin 60º = \frac{BC}{8} \]
they tell you sin 60 is about 0.866
\[ 0.866 = \frac{BC}{8} \]
to find BC, multiply both sides by 8, like this
\[ 0.866 \cdot 8 = \frac{BC}{\cancel{8}} \cdot \cancel{8}\]
simplify to find BC