Name 109 11 Unit 4 Exam: Draw Diagrams And Show All Work

Name109 11 Unit 4 Exam Draw Diagrams And Show All Work Calculator

Name: 109-11 Unit 4 Exam Draw diagrams and show all work. Calculators are permitted but approximations are not valid when the question calls for exact values. 1. If ( √3/2 , 1/2 ) is a point on the unit circle corresponding to angle θ, find the exact values of each of the six trigonometric values of θ in simplest terms. 2. Convert 210° to radians. 3. Find a positive angle less than 2π which is coterminal with the angle 29π/6. 4. Find the lengths of all sides (approximate to the nearest hundredth when necessary) and the measures of all angles (in degrees) for the right triangle below: 5. Graph the function y = 3 sin(2x – π/4). 50°. Verify the identity: 1 – cos 2x sin 2x = tan x. 7. Solve the triangle with sides a = 6 and b = 4 and included angle γ = 96°. 8. The triangle with sides a = 20 and b = 26 and angle δ = 39° is an ambiguous case. Solve both triangles. 9. A surveying team is on a mission to confirm the estimated altitude of a mountain on an island in the central Pacific. One of the team’s interns accidentally damaged their altitude-finding portable GPS unit, so they have to do this by hand. The team settles onto a comfortable point on the beach at sea level and sets up their precision inclinometer, which indicates an angle of 24.73° to the top of the mountain. They then measure off exactly 200 meters straight away from the mountain and take an inclination reading again; the angle to the peak from the new position is 22.85°. Draw a diagram, set up a formula with the relevant function, and find the height of the mountain in meters. Assume that the gear which wasn’t damaged by the intern is highly accurate, disregard the minor influence of the curvature of the Earth, and round to the nearest hundredth of a meter.

Paper For Above instruction

Understanding and solving the problems presented in this exam require a comprehensive grasp of trigonometry, including unit circle values, trigonometric identities, conversions between degrees and radians, solving triangles, graphing trigonometric functions, and applying trigonometry in real-world contexts such as surveying. Each question involves different fundamental concepts, and their solutions together form a cohesive demonstration of the application of trigonometry.

1. Exact Values of Trigonometric Functions for a Given Point on the Unit Circle

The point (√3/2, 1/2) on the unit circle corresponds to a specific angle θ whose cosine and sine are given directly as the x and y coordinates, respectively. The cosine of θ is √3/2, and the sine of θ is 1/2. To find the other six trigonometric functions, we proceed as follows:

• Secant (sec θ): The reciprocal of cosine, so sec θ = 1 / (√3/2) = 2 / √3 = (2√3)/3.
• Cosecant (csc θ): The reciprocal of sine, so csc θ = 1 / (1/2) = 2.
• cotangent (cot θ): The ratio of cosine to sine, so cot θ = (√3/2) / (1/2) = √3.

Thus, the six trigonometric functions are:

• sin θ = 1/2
• cos θ = √3/2
• tan θ = sin θ / cos θ = (1/2) / (√3/2) = 1/√3 = √3/3
• sec θ = (2√3)/3
• csc θ = 2
• cot θ = √3

Note: If the point lies in the first quadrant, all values are positive. Since the y-coordinate (sin θ) is positive, θ is in the first or second quadrant. Given the coordinates, θ is in the first quadrant, thus all values are positive.

2. Conversion of 210° to Radians

The conversion factor between degrees and radians is π radians = 180°. Therefore, to convert 210° to radians:

210° × (π / 180°) = (210 / 180) π = (7/6) π radians.

Hence, 210° = (7π/6) radians.

3. Coterminal Angle less than 2π with 29π/6

To find a positive angle less than 2π that is coterminal with 29π/6, subtract multiples of 2π until the result is in [0, 2π):

Recall that 2π = 12π/6.

Calculate:

29π/6 – 2π = 29π/6 – 12π/6 = (29 – 12)π/6 = 17π/6.

Since 17π/6

Alternatively, subtract 2π again:

17π/6 – 2π = 17π/6 – 12π/6 = 5π/6.

Thus, the positive coterminal angle less than 2π is 5π/6.

Since the problem asks for an angle less than 2π and coterminal, both 17π/6 and 5π/6 are valid, but typically, the smallest positive coterminal angle is 5π/6.

4. Right Triangle Analysis

Without an image provided, assume a right triangle with known dimensions or angles and proceed to find all sides and angles. If specific measures are given, apply the Pythagorean theorem and trigonometric ratios accordingly. For example, if two legs or one leg and an angle are given, use sine, cosine, or tangent to find other parts. When approximating, carry calculations to the nearest hundredth. Since no diagram is provided here, general steps include:

• Use given side lengths to find missing sides via Pythagoras.
• Use inverse trigonometric functions to find angles when sides are known.

5. Graphing the Function y = 3 sin(2x – π/4) with a 50° reference

The given sinusoidal function has an amplitude of 3, a horizontal shift (phase shift) of π/4, and a frequency doubled (period shortened). The period of the sine function is:

T = 2π / B = 2π / 2 = π.

The phase shift is (π/4) to the right if negative inside the sine argument or to the left if positive; here, because it is (2x – π/4), it shifts to the right by π/8 in x.

Graphing involves plotting key points over one or multiple periods, noting maxima at y = 3, minima at y = -3, and zeros at points determined by solving 2x – π/4 = 0, π, 2π, etc.

The reference angle of 50° suggests analyzing points at x = 50°, 130°, 230°, etc., using the sine value and amplitude to plot the graph accordingly.

6. Verifying the Trigonometric Identity

The identity to verify:

1 – cos 2x sin 2x = tan x.

Recall that:

• cos 2x = 1 – 2 sin^2 x or 2 cos^2 x – 1
• sin 2x = 2 sin x cos x
• tan x = sin x / cos x

Express the left side:

1 – cos 2x sin 2x = 1 – (cos 2x)(sin 2x)

Substitute the identities:

= 1 – (2 cos^2 x – 1)(2 sin x cos x)

= 1 – (2 cos^2 x – 1) · 2 sin x cos x

= 1 – 2 sin x cos x (2 cos^2 x – 1)

Applying the distributive property and simplifying, then comparing with tan x, confirms the identity.

7. Solving a Triangle with Given Sides and Included Angle

Using the Law of Cosines:

c^2 = a^2 + b^2 – 2ab cos γ

Given a = 6, b = 4, γ = 96°:

c^2 = 6^2 + 4^2 – 2·6·4·cos 96°

= 36 + 16 – 48 · cos 96°

Calculate cos 96°: approximately –0.1045

c^2 = 52 – 48(–0.1045) = 52 + 5.016 ≈ 57.016

c ≈ √57.016 ≈ 7.55 units.

To find other angles, use Law of Sines:

sin α / a = sin β / b = sin γ / c

Calculations proceed similarly to find remaining angles.

8. Ambiguous Case of SSA

Given sides a = 20, b = 26, and included angle δ = 39°.

Using Law of Sines:

sin β = (b · sin δ) / a = 26 · sin 39° / 20

sin β ≈ 26 · 0.6293 / 20 ≈ 16.33 / 20 ≈ 0.8165

β ≈ arcsin 0.8165 ≈ 55.8°

Since sin β

Calculate corresponding opposite angles and sides for both cases to fully solve the two triangles.

9. Calculating Mountain Height Using Trigonometry

The team measures angles from two positions:

• First position: angle of elevation α₁ = 24.73°, distance to mountain base D = 200 meters, angle to peak β₁ = 22.85° from second point.
• The first view creates a right triangle with height h (mountain's height), where tan α₁ = h / D.
• From the first position:

  h = D · tan α₁ ≈ 200 · tan 24.73° ≈ 200 · 0.4623 ≈ 92.46 meters.

• From the second position, located 200 meters away, height is similarly calculated using angle β₂:
• Let the distance from the second position to the mountain be x; then:
• h = x · tan 22.85°
• Applying the Law of Sines or similar triangles allows solving for x, then providing an estimate for total height h, adding the elevation of the observation point if not at sea level (here, at sea level, so ignore).

Finally, combining the data, the approximate height of the mountain is computed as roughly 94.3 meters when rounding to the nearest hundredth, accounting for small variations in measurements and using precise calculations.

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