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?
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@ranga @RosieF @robtobey
@Johnbc
Do you know the trigonometric functions for angles and sides of the angles?
sines and cosines?
Yup
Do you know the sides that Sine, Cosine, and tangent relate to?
not really..
i know the formulas for sines and cosines
What are they?
for sines:\[\frac{ SinA }{ a }=\frac{ SinB }{ b }=\frac{ SinC }{ c }\]
Cosines:\[a^2=b^2+c^2-2bc*\cos(A)\]
\[b^2=a^2+c^2-2ac*\cos(B)\]
\[c^2=a^2+b^2-2ab*\cos(C)\]
those are the formulas
Those are definitely the formulas but there are easier forms for angles
there is?
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okay
im looking for all 3 angles
Yup
And depending on the angle that you want to find first you can use the corresponding formula
okay so its CAH. Adjacent and Hypotenuse
This is NOT a right triangle. |dw:1400442844816:dw|
Use the Law of Cosines to find angle A. a^2 = b^2 + c^2 - 2bc * cos(A) 15^2 = 20^2 + 12^2 - (2)(20)(12)cos(A). Solve for angle A. Then you can use the Law of Sines to find angle B a / sin(A) = b / sin(B). Then you can find angle C as 180 - angle A - angle B
im not doing something right, i got 0.84
15^2 = 20^2 + 12^2 - (2)(20)(12)cos(A). 225 = 400 + 144 - 480cos(A) 480cos(A) = 400 + 144 - 225 = 319 cos(A) = 319/480 = 0.6645833 A = arccos(0.664583) = 48.35 degrees.
okay so m<A= 48.35degress
Find angle B using the Law of Sines.
94.94 degrees
I have not done the calculation but I will have to assume you did it correctly.
Yes, 94.94 degrees is correct.
the other angle is 36.71
Yes. Not sure at what decimal place they want you to round off the angles.
2
okay, then you have the answer.
okay thank you so much
You are welcome.
the answer is 48.35°, 94.94°, 36.71°
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