Name:(printed)
Signature:
GUID:
Section:
KEY KEY KEY
MA110 Exam 3: Fall 2015
Sections 007-012
Directions
This exam consists of this cover sheet (page1), five (5) pages of problems and questions (pages
2-6), and two blank pages (numbers 7 and 8) for additional work space. There are a total 10
questions (some with multiple parts). Each question is worth a total of 10 points.
Each question is followed by space to write your answer. Please write your solutions neatly
in this space. There should be ample room for your answer but if you need more space then
you can continue your responses on pages 7 and 8. Be sure to direct the reader to work on
those pages and clearly identify there what goes with any particular question. The graders will
not look on pages 7 and 8 unless you specifically direct them to work there. The backs of
the exam pages will not be visible to the graders.
1. No books or notes may be used on this exam.
2. Turn off your cell phones and put them away until you have turned in your
exam and left the exam room. In view of this instruction, any use of a cell phone
during the exam will be considered cheating.
3. Once the exam has started you may NOT use any entertainment or communication
device in the exam room, including earbuds, headphones, bluetooth devices , etc.
4. Please use a dark pencil or pen.
5. Clearly indicate the reasoning or calculations used to arrive at your answers. Keep in
mind that (except for question 1) your work, justifications, and explanations are what
will be graded. (Unsupported answers, including answers simply taken from a calculator,
will receive no credit).
6. Be sure to erase or cross out anything you do not want graded. Anything in the response
area that is not clearly crossed out or erased will be considered part of the intended
answer.
7. You may not use a computer on this exam. You may use a calculator, but not one which
has symbolic manipulation capabilities. In addition, answers that are simply taken from
a calculator and presented without proper justification will receive no credit.
Some possibly useful formulas are at the bottom page number 6.
1
Last Name:(printed)
GUID:
(1) (10 points) Complete the following to statements of trig identities:
(a) sin(A + B) =
sin(A) cos(B) + cos(A) sin(B)
(b) cos(A + B) =
cos(A) cos(B) − sin(A) sin(B)
(c) sin(−A) =
− sin(A)
(d) cos(−A) =
cos(A)
(e) sin(π − A) =
(f) cos( π
2 − A) =
sin(A)
sin(A)
(2) (10 points) Suppose that the locator point of A is (.7660, .6427) and the locator point of
B is (.2588, .9659). Use the appropriate identities to complete the following table of
locator points. Express your answers to 4 decimal places. You must provide work that
makes it clear how you arrive at your answers. Calculators may be used for the arithmetic.
angle
locator point
A:
( .7660 , .6427 )
B:
( .2588 , .9659 )
− .42254313 , .90621016 )
A+B :
(
A−B :
( .81902473 ,
− .57354864 )
2A :
( .17369271 , .98461640 )
B :
2
( .7933473388 , .6087692502 )
2
Last Name:(printed)
GUID:
(3) (10 points) Answers to parts (a) and (c) can be in any format presented in this course.
All answers for this problem must be exact.
(a) What is the equation of the circle of radius 4 which has its center at (−2, 5)?
(x + 2)2 + (y − 5)2 = 42 , (x2 + y 2 + 4x − 10y + 13 = 0)
(b) What are the center and radius of the circle which has equation
x2 + y 2 + 8x − 6y + 21 = 0?
Show your work.
−8
2
= −4,
−(−6)
2
= 3, (−4)2 + 32 − 21 = 4 = 22
center =(−4, 3) , radius = 2.
(c) If A = (2, 3) and B = (1, 5), what is the equation of the circle for which the line
segment AB is a diameter? Show your work.
(x − 2)(x − 1) + (y − 3)(y − 5) or x2 + y 2 − 3x − 8y + 17 or
(x − 23 )2 + (y − 4)2 = 5/4
(4) (10 points) The clock reads 6 : 50. What exactly is the radian measure of the angle
between the two clock hands? Show your work.
10 π
( 10
12 )2π − (6 + 12 ) 6 =
19π
36
3
Last Name:(printed)
GUID:
(5) (10 points) Let C1 be the circle which has equation (x + 3)2 + (y − 1)2 = 9 and the circle
C2 which has equation (x + 1)2 + (y + 1)2 = 5.
(a) What is the equation of the line which passes through the points of intersection of C1
and C2 ?
2
2
2
2
((x + 3) + (y − 1) = 9) − ((x + 1) + (y + 1) = 5)
4x − 4y + 4 = 0
y =x+1
(b) What are the points of intersection of C1 and C2 ? Show your work.
2
2
(x + 1) + ((x + 1) + 1) = 5
2x2 + 6x = 2(x)(x + 3), x = 0, x = −3
(0, 0 + 1) = (0, 1) and (−3, −3 + 1) = (−3, −2)
alt: ((0 + 3)2 + (y − 1)2 = 9, y = 1
(−3 + 3)2 + (y − 1)2 = 9, y = 1 ± 3 = 4 or − 2
must check that 4 is extraneous by checking that
(−3, 4) is not on both circles.
(6) (10 points) If ∠A measures 40 degrees, b = 10 inches and c = 4 inches.
What is the length of a? Show your work.
a2 = b2 + c2 − 2bc cos(40o) = 102 + 42 − 2(10)(4)(.7660)
a2 = 54.72, a = 7.828 in.
4
Last Name:(printed)
GUID:
(7) (10 points) In the diagram, the angle ∠A measures 20 degrees, ∠C measures 50 degrees,
and a, the length of CB is 6 inches. What are b and c, the lengths of AC and AB? Show
your work.
B = 180 − (20 + 50) = 110
sin(C)
sin(B)
sin(A)
=
=
c
b
a
sin(110o )
sin(50o )
sin(20o )
=
=
6
b
c
.34202
.93969
.7660
=
=
6
b
a
c = 13.44, b = 16.48 in.
(8) (10 points) The angle ∠ACB in the diagram measures 130 degrees and the circle is centered at C and has a radius of 10 inches. What is the area of the triangle 4ACB? Show
your work.
Area = area of two half-triangles =
o
130o )) =
2( 12 (10 cos( 130
)(10
sin(
2
2
100(.4226)(.9063) = 38.302 square inches
5
Last Name:(printed)
GUID:
(9) (10 points) From the point A, the tall tree at B ( represented by the segment BC) subtends an angle of π
6 radians while from point M which is 150 feet closer it subtends an
π
angle of 3 radians. What is the height of the tree?
(Note that the diagram is not to scale. Triangle 4ABC is a right triangle with hypotenuse
AC.) Show your work.
sin(30o )
150
=
√
sin(120)
AC
=
√
3
2
AC
√
√
BC
√ , BC = 75 3
= AC , AC = 150 3. 12 = sin(30o ) = BC
, 1 = 150
AC 2
3
alternate solution: ∠AM C = 180 − 60 − 120 degrees so ∠ACM = 30 degrees
so triangle
√
BC
= BC
4AM C is√ isosceles. Therefore AM = M C so M C = 150 so sin(60o ) = 23 = M
C
150
3
so BC = 2 150 feet. alternate solution
BC
tan(A) = 150+M
B
BC
tan( π3 ) = √13 = 150+M
B
√
(150 + M B) = 3BC
BC
tan( π6 ) = tan(M ) = M
B
√
BC
3= M
B
√
3M B = BC, M B = √13 BC
√
(150 + √13 BC) = 3BC
√
150
BC = √3−
= 75 3 = 129.903 feet.
√1
1
2
3
2
150
3
(10) (10 points) Line L has equation y = 2x + 1 and line M has equation y = x. What are the
points on line M whose distance from line L is exactly 3? Show your work.
√|2x−y+1|
=3
2
2
2 +(−1)
on M ( x = y) so √|2x−x+1|
=3
22 +(−1)2
√
|x + 1| = 3 √5
x + 1 = ±3 √5
√
x = −1 √
− 3 5, x √
= −1 + 3 5
√
(−1 − 3 5, −1 − 3 5) and (−1 + 3 5, −1 + 3sqrt5)
(−3.236067977, −3.236067977) and (1.236067977, 1.236067977)
6
The following formulas may be useful:
r
1 + cos(θ)
|ax + by + c|
θ
tan(a) + tan(b)
d= √
, cos( ) = ±
, tan(a + b) =
1 − tan(a) tan(b)
2
2
a2 + b 2
7
Last Name:(printed)
GUID:
This page is provided in case you need additional space. If work on this page is part of
your solution to a problem then you must make a note on the space under the problem to look
here and you must clearly indicate what on this page is part of your solution to any particular
problem.
8
Last Name:(printed)
GUID:
This page is provided in case you need additional space. If work on this page is part of
your solution to a problem then you must make a note on the space under the problem to look
here and you must clearly indicate what on this page is part of your solution to any particular
problem.
9