15. a ship at sea, the hmss lonestar, spots two other ships, the hmss newsome and the ss parry, and measures…

15. a ship at sea, the hmss lonestar, spots two other ships, the hmss newsome and the ss parry, and measures the angle between them to be 27°. the distance between the lonestar and the newsome is 199m. there is an angle of 52° between the lonestar and the parry. what is the distance between the newsome and the parry, to the nearest meter?\n16. for each triangle, calculate the value of w to the nearest tenth of a metre.\na)\n!triangle abc(https://example.com/abc.png)\n60.0 m\n49° 53°\nb d c\nw\n\nb)\n!triangle ghj(https://example.com/ghj.png)\n55 m\n40° 45°\ng h j\nw\n\n17. complete the chart below.\n| function | amplitude | period | phase shift (with direction) | vertical displacement | maximum | minimum |\n| ---- | ---- | ---- | ---- | ---- | ---- | ---- |\n| (y = 2sin\theta+5) | | | | | | |\n| (y=\frac{1}{4}cos(2\theta - 60^{circ})) | | | | | | |\n| (y = 5cos3(\theta - 45^{circ})-7) | | | | | | |\n\n18. list the domain and range for each of the transformed functions in question 17.\n19. describe the transformations and then graph both the base - curve and the transformed curve.\na) (y = 2cos(\theta + 30^{circ})+1)\nb) (y=\frac{1}{2}sinleft\frac{1}{3}(\theta - 20^{circ})\right-2)\n20. create an equation for each of the functions described.\na) period (120^{circ}), amplitude 2 and max ((45^{circ},1)) given (y = sin\theta)\nb) the maximum is 3 and the minimum is - 3. the period is (180^{circ}) and the graph begins at ((0,0)) given (y=cos\theta)\n21. prove that\na) (\tan x+\frac{1}{\tan x}=\frac{1}{sin xcos x})\nb) (\frac{1 + cos\theta}{sin\theta}+\frac{sin\theta}{1+cos\theta}=\frac{2}{sin\theta})\nc) (2sin\theta(1 + cos\theta)=\frac{2sin^{3}\theta}{1-cos\theta})\nd) (\frac{1}{\tan\theta}+cos\theta=\frac{cos\theta(1+sin\theta)}{sin\theta})

15. a ship at sea, the hmss lonestar, spots two other ships, the hmss newsome and the ss parry, and measures the angle between them to be 27°. the distance between the lonestar and the newsome is 199m. there is an angle of 52° between the lonestar and the parry. what is the distance between the newsome and the parry, to the nearest meter?\n16. for each triangle, calculate the value of w to the nearest tenth of a metre.\na)\n!triangle abc(https://example.com/abc.png)\n60.0 m\n49° 53°\nb d c\nw\n\nb)\n!triangle ghj(https://example.com/ghj.png)\n55 m\n40° 45°\ng h j\nw\n\n17. complete the chart below.\n| function | amplitude | period | phase shift (with direction) | vertical displacement | maximum | minimum |\n| ---- | ---- | ---- | ---- | ---- | ---- | ---- |\n| (y = 2sin\theta+5) | | | | | | |\n| (y=\frac{1}{4}cos(2\theta - 60^{circ})) | | | | | | |\n| (y = 5cos3(\theta - 45^{circ})-7) | | | | | | |\n\n18. list the domain and range for each of the transformed functions in question 17.\n19. describe the transformations and then graph both the base - curve and the transformed curve.\na) (y = 2cos(\theta + 30^{circ})+1)\nb) (y=\frac{1}{2}sinleft\frac{1}{3}(\theta - 20^{circ})\right-2)\n20. create an equation for each of the functions described.\na) period (120^{circ}), amplitude 2 and max ((45^{circ},1)) given (y = sin\theta)\nb) the maximum is 3 and the minimum is - 3. the period is (180^{circ}) and the graph begins at ((0,0)) given (y=cos\theta)\n21. prove that\na) (\tan x+\frac{1}{\tan x}=\frac{1}{sin xcos x})\nb) (\frac{1 + cos\theta}{sin\theta}+\frac{sin\theta}{1+cos\theta}=\frac{2}{sin\theta})\nc) (2sin\theta(1 + cos\theta)=\frac{2sin^{3}\theta}{1-cos\theta})\nd) (\frac{1}{\tan\theta}+cos\theta=\frac{cos\theta(1+sin\theta)}{sin\theta})

Answer

Explanation:

Step1: Identify relevant trig - concepts

For 15, use the Law of Sines. For 16, use right - triangle trigonometry. For 17 - 21, use trigonometric function properties.

Step2: Solve problem 15

Let (A = 27^{\circ}), (B = 52^{\circ}), (c = 199)m. First find (C=180-(27 + 52)=101^{\circ}). By the Law of Sines (\frac{a}{\sin A}=\frac{c}{\sin C}), so (a=\frac{c\sin A}{\sin C}=\frac{199\sin27^{\circ}}{\sin101^{\circ}}\approx91)m.

Step3: Solve problem 16a

In (\triangle ABD), (\tan49^{\circ}=\frac{AD}{BD}), in (\triangle ACD), (\tan53^{\circ}=\frac{AD}{CD}). Let (AD = h), (BD=\frac{h}{\tan49^{\circ}}), (CD=\frac{h}{\tan53^{\circ}}), (w = BD + CD), (h = 60)m. (w=60(\frac{1}{\tan49^{\circ}}+\frac{1}{\tan53^{\circ}})\approx60(0.869 + 0.754)\approx97.4)m.

Step4: Solve problem 16b

In right - triangle (GJH), (\tan40^{\circ}=\frac{HJ}{GH}), (HJ = 55)m, (w=GH=\frac{55}{\tan40^{\circ}}\approx65.5)m.

Step5: Solve problem 17

For (y = 2\sin\theta+5): Amplitude (= 2), Period (=360^{\circ}), Phase Shift (=0), Vertical Displacement (=5), Maximum (=7), Minimum (=3). For (y=\frac{1}{4}\cos(2\theta - 60^{\circ})): Amplitude (=\frac{1}{4}), Period (=180^{\circ}), Phase Shift (=30^{\circ}) right, Vertical Displacement (=0), Maximum (=\frac{1}{4}), Minimum (-\frac{1}{4}). For (y = 5\cos3(\theta - 45^{\circ})-7): Amplitude (=5), Period (=120^{\circ}), Phase Shift (=45^{\circ}) right, Vertical Displacement (-7), Maximum (- 2), Minimum (-12).

Step6: Solve problem 18

For (y = 2\sin\theta+5): Domain (\theta\in R), Range (y\in[3,7]). For (y=\frac{1}{4}\cos(2\theta - 60^{\circ})): Domain (\theta\in R), Range (y\in[-\frac{1}{4},\frac{1}{4}]). For (y = 5\cos3(\theta - 4[Client Connection Error]