e system of equations
tan(60°) = x/d and
tan(30°) = x/(d + 1000).
Solving for x
yields x = 1000(tan(30°))/(1  cot(60°)tan(30°)).
Alternatively,
you could recognize that, here, the hypotenuse of the triangle
created by the 60° sighting is one of the two equal sides of the
isosceles 30°,30°,120° triangle formed from the 30° sighting.
Thus, that hypotenuse has length 1000 feet. This leads to the
very simple equation x = 1000 sin(60°).
In either case, you will
get x
≈
866.03 feet and so h
≈
871.03 feet.
[Warning: The cute
isosceles triangle thingy doesn’t usually happen.
The other
approach is more general.]
14. (5 pts.)
If the polar coordinates of a point are given by
(r,
θ
) = (9.5,110°), find the rectangular coordinates for the
point.
In doing this, make clear which values are exact and
which are approximations.
(x,y) = ((9.5)cos(110°),(9.5)sin(110°))
≈
(3.25, 8.93)
15. (5 pts.)
If the rectangular coordinates of a point are
given by (x,y) = (5,5
√
3), obtain polar coordinates for the
point.
It’s easy to see r
2
= 100.
Thus, use r = 10, to keep things
simple.
It turns out that the reference angle
θ
r
satisfies the
equation tan(
θ
r
)=3
1/2
.
Thus, because the point lies in the
third quadrant, we may use either
θ
= 240° or
θ
=4
π
/3.
From
here, it is easy to list all pairs (r,
θ
) that represent the
point.
16. (10 pts.)
(a) Obtain all solutions to the equation below,
and then (b) list the solutions
θ
with 0
≤θ
<2
π
.
2 sin
2
(
θ
) + 3 sin(
θ
)+1=0
(a)
The given equation is equivalent to
(2sin(
θ
) + 1)(sin(
θ
)+1
)=0
,
which is equivalent to sin(
θ
) = 1/2 or sin(
θ
) = 1.
All
solutions to sin(
θ
) = 1/2 are given by
θ
=7
π
/
6+2
k
π
or
θ
=1
1
π
/
6+2
k
π
, k any integer, and all solutions to sin(
θ
)=
1
are given by
θ
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 Spring '08
 Storfer
 Pythagorean Theorem, Sin, pts, triangle

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