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How do you solve the #triangle RST# given #R=35^circ, s=16, t=9#?

Answer 1

You're given the included angle. Use the Law of Cosines to solve this problem. The formula : #a^2=b^2+c^2-2bc*cosA# Let's replace #a# with #r#, #b# with #s#, and #c# with #t#. Let's solve for side #r# first. Substitute values. #r^2=16^2+9^2-(2*16*9)*cos(35)# #r^2=256+81-288*cos(35)# #r^2approx337-235.92# #r^2approx101.09# #r approx 10.05#

Now, let's solve for angle S. You can solve using the Law of Sines, but I'll just use the Law of Cosines because you didn't seem to know how to do it.

Also, for the sake of simplicity, I'll just use #r=10# The rearranged formula of the Law of Cosines (and changed to our variables r, s and t) : #cos S = (r^2+t^2-s^2)/(2rt)# Substitute values : # cos S = (10^2+9^2-16^2)/(2*16*9)# #cos S = (100+81-256)/288# # cos S = -25/96# #S=cos^-1 (-25/96)# #S approx 74.91#

You can work out the remaining angle using the angles in a triangle add up to 180 rule.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.