Solve the problem. This problem is very difficult to understand. Let s see if we can make sense of it. Note that there are multiple interpretations of the problem and that they are all unsatisfactory. The problem does not say the lake is a circle, but if it is not, the problem cannot be solved. So, let s assume the lake is a circle. What is a transit? From www.surveyhistory.org, we get The transit is used by the surveyor to measure both horizontal and vertical angles. A transit A piling What is a piling? From wisegeek.com, a piling is a component in a foundation which is driven into the ground to ensure that the foundation is deep. We will need a right angle to solve a problem with only one length and angle given, so let s infer a 90 angle as shown in the following diagram. Based on the diagram, which meets every criterion laid out in the problem, we now have: tan 35 90 90 tan 35 ft. The body of water in this problem would be better described as a pond than a lake! Page 1 of 10
Based on the illustration at right, we get the following: tan 90 200.45 tan.45 The key on this type of problem is to draw the correct triangle. Recall that tan Since is in 2, we can plot the point 3, 8 to help us. Notice that the hypotenuse must have length: 3 8 73. Then, cos The angle is in Q2, but tangent is defined only in Q1 and Q4. Further, tan 0 in Q2. So we seek the angle in Q4, where tangent is also 0, with the same tangent value as. Recall that the tangent function has a period of radians. Then, tan tan Page 2 of 10
5) Find the exact value of the expression. a. cos b. sin c. tan Know where the primary values for the inverse trig functions are defined. sin θ is defined in Q1 and Q4. cos θ is defined in Q1 and Q2. tan θ is defined in Q1 and Q4. You can think of this as adding two vectors with the bearings identified in the problem. Step 1: Convert each vector into its i and j components, using reference angles. Let u be a vector of 25 mi. at bearing: N63 W From the diagram at right, θ 90 63 27 25 cos 27 22.2752 25 sin 27 11.3498 Let v be a vector of 24 mi. at bearing: S27 W From the diagram at right, φ 90 27 63 24 cos63 10.8958 24 sin63 21.3842 Step 2: Add the results for the two vectors 22.2752, 11.3498 10.8958, 21.3842 33.1710, 10.0344 Step 3: Find the resulting angle and convert it to its bearing form (see diagram above right). tan 10.0344 33.1710 16.8 Bearing S90 16.8 W. Page 3 of 10
7) A ship is 50 miles west and 31 miles south of a harbor. What bearing should the Captain set to sail directly to harbor? θ tan 31.8 φ 90 31.8 58.2. For Problems 8 and 9, use: or with 0 Harmonic motion equations: cos or sin The spring will start at a value of 8, and oscillate from 8 to 8 over time. A good representation of this would be a cosine curve with lead coefficient 8, because cos0 1, whereas, sin0 0. The period of our function is 5 seconds. So, we get: 1 period 1 5 2 2 1 5 2 5 The resulting equation, then, is: Note: both of these functions can be graphed using the Trigonometry app available at www.mathguy.us. Assuming that 0 at time 0, it makes sense to use a sine function for this problem. Since the amplitude is 3 feet, a good representation of this would be a sine curve with lead coefficient 3, because sin0 0, whereas, cos0 0. Recalling that 2, and with we get: 2 5. The resulting equation, then, is: Page 4 of 10
cos sin 1 tan cos sin cos cos sin cos cos sin cos sin cos cos sin 1 cos cos sin sin sin cot sin sin cos sin sin cos Answer B sin 75 sin 1 2 150 1cos150 2 1 3 2 2 Note: the half angle formula has a "" sign in front. We use for this problem because 75 is in Q1, and sine values are positive in Q1. cos 3 8 cos1 2 3 4 1cos3 4 2 1 2 2 2 Note: the half angle formula has a "" sign in front. We use for this problem because is in Q1, and cosine values are positive in Q1. Page 5 of 10
tan sintan tan sin tan 0 tan sin 1 0 tan 0 or sin 1 0 0,π sin 1, While is a solution to the equation sin 1, tan is undefined at. Therefore, the original equation, tan sintan is undefined at. So is not a solution to this equation. 2 cos sin20 2 1 sin sin20 22sin sin20 2 sin sin0 sin 2 sin 1 0 When an equation contains more than one function, try to convert it to one that contains only one function. sin 0 or 2 sin 1 0 0,π sin,,,, Page 6 of 10
cos 0 180 a 1, 1 4, 5 1 4 1 5 1 1 1 2 4 5 41 cos cos 1. 82 b 1, 2 3, 3 1 3 2 3 9 1 2 5 3 3 3 2 cos cos 3. 10 Page 7 of 10
Substitute cos and 3 3 3 3 0 3 9 4 9 4 Substitute xr cos Substitute cos,sin and 4 8 48 48 4 8 0 44 816 416 Page 8 of 10
a) The process of putting a complex number in polar form is very similar to converting a set of rectangular coordinates to polar coordinates. So, if this process seems familiar, that s because it is. 33 3 3 3 3 6 tan 3 3 3 tan 3 in 2 2 3 So, the polar form is: 6 cos sin b) 5 3 5 5 3 5 10 tan 5 5 3 tan 3 3 in 3 7 6 So, the polar form is: 10 cos sin 3 cos 15 sin 15 To take a power of two numbers in polar form, take the power of the value and multiply the angle by the exponent. (This is the essence of DeMoivre s Theorem.) 3 cos 15 sin 15 3 cis4 15 81 cis60 81 cis60 81 cos 60 sin 60 81 1 2 3 2 Page 9 of 10
3 3 3 3 3 3 6 tan 3 3 3 tan 3 in 2 2 3 Then, 3 3 3 6 cis 6 cis 3 216 cis 2 216 cos 2 sin 2 Page 10 of 10