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Quiz 9
Math 1151, Precalculus 2
Name:
Summer 2012, Section 002
Problem 1a.(5 points) What is the law of cosines?
c2 = a2 + b2 − 2ab cos(C)
Problem 1b.(5 points) What is the law of sines?
sin A
a
=
sin B
b
=
sin C
c
Problem 2.(10 points) Solve the triangle which has sides of lengths a=9, b=7, c=10.
(Round to the nearest tenth)
This is a SSS triangle so we must use the law of cosines to start solving it.
2 −a2 −b2
C = cos−1 ( c
−2ab
2
2
2
−7 −9
) = cos−1 ( 10−2(9)(7)
) = cos−1 (5/21) = 76.2◦ .
2 −c2 −b2
We can continue to use the law of cosines to find one more angle: A = cos−1 ( a
2
2
2
−10 −9
cos−1 ( 9−2(10)(7)
) = cos−1 (17/35) = 60.9◦ .
And lastly we find B = 180◦ − 60.9◦ − 76.2◦ = 42.9◦ .
−2cb
)=
Problem 3.(10 points) An airplane is spotted by two people, one at point P and the
other at point Q, who are 1000 feet apart. At one instance each person takes a sighting
of the angle of elevation to the plane as indicated in the illustration. How high is the
plane (find h)?
This is currently a ASA triangle. So we are going to use the law of sines to solve
it. We first start by solving for the last angle which is 180◦ − 85◦ − 70◦ = 25◦ . Now
we can use the law of sines to solve for a or b: b =
1000 sin(70◦ )
sin(25◦ )
1000 sin(85◦ )
sin(25◦ )
= 2357.2 ft and a =
= 2223.5 ft. To find the height of the plane we will use one of the right
triangles: h = b sin(70◦ ) = a sin(85◦ ) = 2215.04 ft.
Extra Credit(10 points) Find a formula for all the vertical asymptotes of cot(2x).
Then list all the vertical asymptotes (in radians) that lie in the interval (−2π, 0).
In order to find the asymptotes we need to solve for when the denominator of cot(2x)
is zero. So we need to solve the trigonometric equation sin(2x) = 0. This occurs when
2x = πk. Solving for x we get an equation that tell us where all the vertical asymptotes:
x = π2 k where k is an integer.