solutions

Math Problem Solutions

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Problem 35 — Photography (Hawk Distance)

Setup: Jeff is taking pictures of a hawk that is 150 ft above him. The hawk will fly directly over Jeff. Let d be the distance from Jeff to the hawk, and θ be the angle of elevation from Jeff’s camera to the hawk.

From the diagram, the hawk is at a height of 150 ft. The relationship between d, θ, and the height forms a right triangle where the opposite side (height) is 150 ft and the hypotenuse is d.

Part (a): Write d as a function of θ

Using the sine ratio:

$\sin (\theta )=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{150}{d}$

Solving for d:

$d=\frac{150}{\sin (\theta )}$

$\boxed{d(\theta )=150\csc (\theta )}$

Part (b): Graph the function on the interval 0 < θ < π

Key features of d(θ) = 150 csc(θ) on (0, π):

• Vertical asymptotes at θ = 0 and θ = π
• Minimum value at θ = π/2, where d = 150 csc(π/2) = 150(1) = 150
• The graph is a U-shaped curve opening upward with minimum at (π/2, 150)
• As θ → 0⁺ or θ → π⁻, d → ∞

θ	d = 150 csc(θ)
π/6	150 csc(π/6) = 150(2) = 300
π/4	150 csc(π/4) = 150(√2) ≈ 212.1
π/3	150 csc(π/3) = 150(2√3/3) ≈ 173.2
π/2	150 csc(π/2) = 150(1) = 150
2π/3	150 csc(2π/3) ≈ 173.2
3π/4	150 csc(3π/4) ≈ 212.1
5π/6	150 csc(5π/6) = 300

(The graph is a csc curve with minimum at 150, symmetric about θ = π/2.)

Part (c): How far away is the hawk when θ = 45°?

$d=150\csc (45°)=\frac{150}{\sin (45°)}=\frac{150}{\frac{\sqrt{2}}{2}}=\frac{150\cdot 2}{\sqrt{2}}=\frac{300}{\sqrt{2}}=\frac{300\sqrt{2}}{2}=150\sqrt{2}$

$\boxed{d\approx 212.1\text{ ft}}$

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Problems 44–47 — Match Each Function With Its Graph

The four graphs (a, b, c, d) need to be matched with these functions:

• 44. y = csc(x/3 + π/4) − 2
• 45. y = sec(x/3 + π/4) − 2
• 46. y = cot(2x − π/4) − 2
• 47. y = tan(2x − π/4) − 2

Analysis of each function:

Problem 44: y = csc(x/3 + π/4) − 2

• This is a cosecant function.
• B = 1/3, so the period = 2π / (1/3) = 6π
• Phase shift = −π/4 ÷ (1/3) = −3π/4 (shift left 3π/4)
• Vertical shift: down 2
• The x-axis scale on graphs (a) and (c) extends to ±4π, which accommodates period 6π.
• Cosecant has U-shaped curves opening up and down.
• Graph (a) shows a csc-type curve with long period and vertical shift down 2.

$\boxed{44\to \text{Graph (a)}}$

Problem 45: y = sec(x/3 + π/4) − 2

• This is a secant function.
• B = 1/3, so the period = 2π / (1/3) = 6π
• Phase shift = −3π/4 (shift left 3π/4)
• Vertical shift: down 2
• Secant has U-shaped curves similar to csc but shifted by half a period relative to csc.
• Graph (c) shows a sec-type curve with long period and vertical shift down 2.

$\boxed{45\to \text{Graph (c)}}$

Problem 46: y = cot(2x − π/4) − 2

• This is a cotangent function.
• B = 2, so the period = π/2
• Phase shift = (π/4)/2 = π/8 (shift right π/8)
• Vertical shift: down 2
• Cotangent is decreasing between consecutive vertical asymptotes.
• Graph (b) shows a cot-type curve (decreasing branches) with short period and vertical shift down 2.

$\boxed{46\to \text{Graph (b)}}$

Problem 47: y = tan(2x − π/4) − 2

• This is a tangent function.
• B = 2, so the period = π/2
• Phase shift = (π/4)/2 = π/8 (shift right π/8)
• Vertical shift: down 2
• Tangent is increasing between consecutive vertical asymptotes.
• Graph (d) shows a tan-type curve (increasing branches) with short period and vertical shift down 2.

$\boxed{47\to \text{Graph (d)}}$

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Problems 52–55 — Write an Equation Given Period, Phase Shift, and Vertical Shift

Problem 52: function: sec; period: 3π; ps: 0; vs: 2

The general form of secant is: y = sec(Bx − C) + D

• Period = 2π/B = 3π → B = 2π/(3π) = 2/3
• Phase shift = C/B = 0 → C = 0
• Vertical shift D = 2

$\boxed{y=\sec \left(\frac{2}{3}x\right)+2}$

Problem 53: function: tan; period: π/2; ps: π/4; vs: −1

The general form of tangent is: y = tan(Bx − C) + D

• Period = π/B = π/2 → B = π/(π/2) = 2
• Phase shift = C/B = π/4 → C = B · (π/4) = 2 · (π/4) = π/2
• Vertical shift D = −1

$\boxed{y=\tan \left(2x-\frac{\pi }{2}\right)-1}$

Problem 54: function: csc; period: π/4; ps: −π; vs: 0

The general form of cosecant is: y = csc(Bx − C) + D

• Period = 2π/B = π/4 → B = 2π/(π/4) = 8
• Phase shift = C/B = −π → C = B · (−π) = 8 · (−π) = −8π
• Vertical shift D = 0

$\boxed{y=\csc (8x+8\pi )}$

Note: Since csc has period 2π in its argument, and 8·(−π) shifts by a full multiple of the internal period, this simplifies to y = csc(8x). However, the non-simplified form showing the phase shift is:

$y=\csc \left(8x+8\pi \right)=\csc (8x)$

$\boxed{y=\csc (8x+8\pi )\text{ or equivalently }y=\csc (8x)}$

Problem 55: function: cot; period: 3π; ps: π/2; vs: 4

The general form of cotangent is: y = cot(Bx − C) + D

• Period = π/B = 3π → B = π/(3π) = 1/3
• Phase shift = C/B = π/2 → C = B · (π/2) = (1/3)(π/2) = π/6
• Vertical shift D = 4

$\boxed{y=\cot \left(\frac{1}{3}x-\frac{\pi }{6}\right)+4}$

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Problem 76 — Right Triangle: Find b given c = 14 and a 30° angle

From the figure, we have a right triangle with:

• Hypotenuse c = 14
• Angle = 30°
• Side b is adjacent to the 30° angle (the base)

Using cosine:

$\cos (30°)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{b}{c}=\frac{b}{14}$

$b=14\cos (30°)=14\cdot \frac{\sqrt{3}}{2}=7\sqrt{3}$

$\boxed{b=7\sqrt{3}\text{(Answer: J)}}$

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Problem 77 — Identify the Equation from the Graph

From the graph:

• The curve has vertical asymptotes at θ = 0 and θ = π/2 (and repeats with period π)
• The curve is decreasing between asymptotes — this identifies it as a cotangent function
• The graph appears to pass through (π/4, 0), meaning the curve is shifted
• Standard cot(θ) has a zero at π/2, but this graph’s zero is at π/4

Checking the options:

• A. y = cot(θ + π/4): Zero when θ + π/4 = π/2 → θ = π/4. Asymptote when θ + π/4 = 0 → θ = −π/4. ✗ (asymptote location doesn’t match)
• B. y = cot(θ − π/4): Zero when θ − π/4 = π/2 → θ = 3π/4. ✗
• C. y = tan(θ + π/4): Tangent is increasing. ✗
• D. y = tan(θ − π/4): Zero when θ − π/4 = 0 → θ = π/4. ✓ Asymptote when θ − π/4 = −π/2 → θ = −π/4 and when θ − π/4 = π/2 → θ = 3π/4.

Wait — re-examining the graph more carefully: the curve passes through approximately (π/4, 0) and appears to be increasing through that zero (going from bottom-left to top-right between asymptotes). That is characteristic of tangent, not cotangent.

The asymptotes appear to be near θ = −π/4 and θ = 3π/4, and the function increases between them, passing through zero at θ = π/4.

For y = tan(θ − π/4):

• Zero at θ = π/4 ✓
• Asymptotes at θ − π/4 = ±π/2 → θ = −π/4 and θ = 3π/4 ✓
• Increasing between asymptotes ✓

$\boxed{\text{D. }y=\tan \left(\theta -\frac{\pi }{4}\right)}$

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Problem 78 — Find θ given sin θ = −1/2 and π < θ < 3π/2

We need to find θ such that:

• sin θ = −1/2
• π < θ < 3π/2 (third quadrant)

The reference angle for sin θ = 1/2 is π/6 (since sin(π/6) = 1/2).

In the third quadrant (π < θ < 3π/2), sine is negative, and:

$\theta =\pi +\frac{\pi }{6}=\frac{6\pi }{6}+\frac{\pi }{6}=\frac{7\pi }{6}$

Verification: sin(7π/6) = sin(π + π/6) = −sin(π/6) = −1/2 ✓

$\boxed{\theta =\frac{7\pi }{6}\text{(Answer: G)}}$

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