Pre-Calculus Help!!...*Please explain how you reached your answer?

Question

A wheel has a diameter of 27.2 inches. What is the length of an arc intercepted by a central angle of 150o? Round your answer to one decimal place.

A. 71.2” B. 35.6” C. 2040” D. 4080”


_____ 2. Let  be an angle of rotation in standard position. If cos  > 0 and csc  < 0, in what quadrant does the terminal side of  lie?

A. Quadrant III B. Quadrant IV C. Quadrant I D. Quadrant II


_____ 3. The point (12, 16) is on the terminal side of angle. Find the exact value of csc .

A. 3/4 B. - 3/4 C. 5/3 D. 5/4
_____4. A building 150 feet casts a 50 foot long shadow. If a person looks down from the top of the building, what is the measure of the angle between the end of the shadow and the vertical side of the building to the nearest degree? Assume the person’s eyes are level with the top of the building.
A. 71o B. 18o C. 72o D. 19o

_____ 5. Find the reference angle for an angle that measures –439o.

A. 101o B. 169o

Answer 1

1. (150 / 360)(2)(3.14)(27.2 / 2) = 35.6 inches
Answer: B

2. cos w > 0 in quadrants I and IV. csc w < 0 in quadrants I and II. So, W is quadrant I.
Answer: C

3. (x, y) = (12, 16). r = sqrt(12^2 + 16^2) = 20.
sin w = y / r
sin w = 16 / 20
sin w = 4 / 5
csc W = 5 / 4
Answer: D

4. tan w = 50 / 150
tan w = 1 / 3
w = arctan (1 / 3)
w = 18.4 degrees
Answer: B

5. -439 + 360 = -79
-79 + 360 = 281
quadrant IV
reference angle = 360 - 281 = 79
?

Answer 2

C= pid = 27.2pi
150/360 *27.2pi = length arc = 35.6 degrees

The cos is > 0 in quadrants 1 and 4
the csc <0 in quadrants 3 and 4
Thus the angle must lie in quadrant IV

csc of angle = 20/16 = 5/4
This is because 20^2 = 12^2 + 16^2 so hypotenuse = 20

arctan (150/50) =arctan 3 = 72 degrees

-439+360 = - 79 degrees
So the reference angle is 79 degrees

Answer 3

1. In this question, you need to determine how big the circumferance of the circle is C = 2pi r or pi d

C = 27.2*pi = 85.45

who big is the arc with a central angle of 150deg, well think of it as the whole circumferance is 360deg so if you set up a ratio between the deg and the length

150deg/360deg = arc/totalcircumferance

150/360 = arc/85.45

arch = (150/360) * 85.45 = 35.6 (answer b)


2. You know that for cosine of any angle to be positive, The terminal side is in quadrants I and IV

for a csc = 1/sin to be negative the terminal side is in quadrants III and IV,

so for both conditions to be true, the terminal side is in IV( answer B)

3. if the terminal side has point (12,16) (you know you are in quadrant 1 (x = 12, y = 16, and r = sqrt(12^2+16^2) = 20) (this is a 3-4-5 triangle) csc(angle) = 1/sin(angle) = 1/(y/r) = r/y = 20/16 = 5/4 (answer D)

4. so the triangle from that angle has the opposite side = 50 and the adjecent side = 150

the tangent of angle looking down = opp/adj so the
the angle looking down = inverse tangent(opp/adj)

= inverse tangent (50/150) = inversetangent(1/3) = 0.333
tan(18) = 0.3249 (closest)
tan(19) = 0.3443

18degrees (answer b)

5. for -439 deg you are going to be in the IV quadrant, If we are using positive angles -439 deg is equivalent to 281 deg to get the reference angle in the IV quadrant subtract the angle from 360

360-281 = 79 (not one of your answers)

Answer 4

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