# [SOLVED] few trigonoetetry questions

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#### Tryptophan

im reviewing for an upcoming test and im having alot of trouble with these problems. any help would be greatly appreciated.

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#### Jhevon

MHF Helper
im reviewing for an upcoming test and im having alot of trouble with these problems. any help would be greatly appreciated.

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I'll do the first and last questions.

For the first question (see figure below).

First, we draw a diagram, which is below. since it is in the fourth quad, the base of the right angled triangle is on the x-axis and it is upside down. also note that all the trig ratios, except cosine and secant, are negative in this quad.

now, cosine is adjacent/hypotenuse, so we have the length of the base and the hypotenuse of the triangle. We use pythagoras' theorem to find that the length of the other side is sqrt(39)...do you need me to show the steps to do this?

now that we know all the sides we can find all the other trig ratios.

sin(x) = opp/hyp = - sqrt(39)/8
csc(x) = 1/sin(x) = - 8/sqrt(39)
sec(x) = 1/cos(x) = 8/5
tan(x) = opp/adj = - sqrt(39)/5
cot(x) = 1/tan(x) = - 5/sqrt(39)

For the second question we use the cosine rule:
the largest angle is the one opposite the largest side:
for reference, i call the side with length 92, AC, the side with length 156 is BC and the side with length 240 is AB. (do you need me to post a diagram for this as well?)

the side opposite to angle BAC is called a, the one opposite angle ACB is called c and the one opposite ABC is called b, this is by convention, it is usually this way.

now to do the work:

By the cosine rule (or some call it the law of cosines)
c^2 = a^2 + b^2 - 2ab*cos(C)...............C is the angle we want to find, it is the angle ACB.
=> c^2 - a^2 - b^2 = -2ab*cos(C)
=> [c^2 - a^2 - b^2]/[-2ab] = cos(C)
=> arccos{[c^2 - a^2 - b^2]/[-2ab]} = C.............arccos means cosine inverse

Now we plug in the values:

=> C = arccos{[240^2 - 156^2 - 92^2]/[-2(156)(92)]}
=> C = arccos{-0.864}
=> C = 149.77 degrees = 2.62 radians

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