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Geometry 16 Online
OpenStudy (anonymous):

To prepare for a triathlon, Amanda starts from position A and rides her bike along a straight road for 12 miles to reach position B. At B, she turns left and rides along another straight road for 15 miles to reach position C. At C, she turns left again and rides 20 miles along a straight road to return to A. In ΔABC, what are m∠A, m∠B, and m∠C, respectively?

OpenStudy (lasttccasey):

Have you learned trigonometry?

OpenStudy (anonymous):

not really

OpenStudy (lasttccasey):

Okay so by m∠A I'm assuming they're asking for the angles?

OpenStudy (anonymous):

yes

OpenStudy (lasttccasey):

Well you will have to use the law of cosines twice then you can find the third easily. So to refresh the law of cosines is: \[\cos A = (b^2 + c^2 - a^2)/2bc\] The best way to do this is draw a triangle and label the sides as lower case with the opposing angle as upper case to easily plug into the formula.

OpenStudy (anonymous):

i get a negative number everytime

OpenStudy (lasttccasey):

You will get one negative, it should be when solving angle B but the angle won't be negative. Are you familiar with this law of cosine? Sorry for the pause.

OpenStudy (anonymous):

The equation is cos A=(15 squared + 20 squared - 12 squared)/2(15)(20) and when i do that i get 72150 and the you cosine it and get a negative .866025404

OpenStudy (lasttccasey):

Ohh I see. Well to solve cos(A) = n you would need to take the inverse. So, \[A = \cos^{-1} (n)\]

OpenStudy (lasttccasey):

For angle A I get cos(A) = (20^2 + 12^2 - 15^2)/(2*20*12) = 0.6645833 Using inverse of cosine, A = 48.3 degrees

OpenStudy (anonymous):

that's what i got too thank you sooo much for your help, I really appreciate it!!!

OpenStudy (lasttccasey):

Good job! It's no problem, for angle B and C I got 94.9 and 36.7 degrees, respectively.

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