Ask your own question, for FREE!
Mathematics 16 Online
OpenStudy (anonymous):

what is the measure of angle C in the triangle below?

OpenStudy (yanasidlinskiy):

Remember when you have "triangle below" make sure you post up the triangle.

OpenStudy (anonymous):

OpenStudy (anonymous):

@YanaSidlinskiy

OpenStudy (anonymous):

@MeganChase please help??

OpenStudy (yanasidlinskiy):

Are you suppose to use the quadratic formula?

OpenStudy (anonymous):

I think so

OpenStudy (anonymous):

The answer choices are A.58.1 B.31.9 C.27.8 D.40.3

OpenStudy (anonymous):

Use the law of Sines: a/sinA =b/SinB=c/SinC Each Lowercase letter is a side Each Uppercase case letter is an angle If you need to find and angle You do a/SinA =b/SinB(unknown angle) A corresponds to a, and so on

OpenStudy (yanasidlinskiy):

Gotta tag this guy again.... @johnweldon1993 My brain isn't working anymore..

OpenStudy (anonymous):

So from my previous reply, you apply it and you get 28/SinC =53/Sin90 Cross multiply: 28*Sin90=53*SinC This all equals to 28/53=SinC Then get rid of Sin from the right side and by doing that you have to multiply it by Sin^-1() on the right AND the left side So Sin^-1(28/53) should be 31.89 degrees

OpenStudy (anonymous):

oh my gosh thank you so much @crazyjulian123

OpenStudy (anonymous):

Do you think you could help me with one more problem? @crazyjulian123

OpenStudy (anonymous):

What is the value of cos(12) to the nearest ten-thousand?

OpenStudy (anonymous):

.9781 it the value to the ten thousands place

OpenStudy (anonymous):

What is the value of sin(42) to the nearest ten-thousand? @MeganChase

OpenStudy (anonymous):

.6691

OpenStudy (anonymous):

Can I ask you one more? its the last one I promise :) Simplify the radical expression. √72x^2 @MeganChase

OpenStudy (anonymous):

ok. wait a second.

OpenStudy (anonymous):

√72x² = √(36•2•x²) = √6²x² = 6x

OpenStudy (anonymous):

is that all?

OpenStudy (anonymous):

Sorry for late response. Use a calculator! It is your best friend for Math!

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!