1. **Solve for the indicated value (right angle triangle):** 1.a) Given $\sin \theta = \frac{4}{7}$, find $\theta$. - Use the inverse sine function: $\theta = \sin^{-1}\left(\frac{4}{7}\right)$. - Calculate $\theta \approx 34.85^\circ$. 1.b) Given $\cos \theta = 0.866$, find $\theta$. - Use the inverse cosine function: $\theta = \cos^{-1}(0.866)$. - Calculate $\theta \approx 30^\circ$. 1.c) Given $\tan \theta = 1.732$, find $\theta$. - Use the inverse tangent function: $\theta = \tan^{-1}(1.732)$. - Calculate $\theta \approx 60^\circ$. 2. **Use SOH-CAH-TOA to solve the right triangles:** 2.a) Opposite = 12, Hypotenuse = 20, find $\theta$. - $\sin \theta = \frac{\text{Opp}}{\text{Hyp}} = \frac{12}{20} = 0.6$. - $\theta = \sin^{-1}(0.6) \approx 36.87^\circ$. 2.b) Adjacent = 9, Opposite = 14, find $\theta$. - $\tan \theta = \frac{\text{Opp}}{\text{Adj}} = \frac{14}{9} \approx 1.555$. - $\theta = \tan^{-1}(1.555) \approx 57.99^\circ$. 2.c) Hypotenuse = 18, Adjacent = 11, find Opposite. - Use Pythagorean theorem: $\text{Opp} = \sqrt{18^2 - 11^2} = \sqrt{324 - 121} = \sqrt{203} \approx 14.25$. 3. **Find the missing side (right angle triangle):** 3.a) Legs: 7 and 24, find Hypotenuse. - $\text{Hyp} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25$. 3.b) Hypotenuse = 17, one leg = 15, find other leg. - Other leg $= \sqrt{17^2 - 15^2} = \sqrt{289 - 225} = \sqrt{64} = 8$. 3.c) Hypotenuse = 20, leg = 12, find other leg. - Other leg $= \sqrt{20^2 - 12^2} = \sqrt{400 - 144} = \sqrt{256} = 16$. 5. **Use Sine Law to solve triangles:** 5.a) $A=42^\circ$, $a=14$, $B=63^\circ$, find $b$. - Law of sines: $\frac{a}{\sin A} = \frac{b}{\sin B}$. - $b = \frac{a \sin B}{\sin A} = \frac{14 \sin 63^\circ}{\sin 42^\circ} \approx \frac{14 \times 0.8910}{0.6691} \approx 18.64$. 5.b) $C=35^\circ$, $c=18$, $B=71^\circ$, find $b$. - $A = 180^\circ - B - C = 180^\circ - 71^\circ - 35^\circ = 74^\circ$. - $\frac{b}{\sin B} = \frac{c}{\sin C} \Rightarrow b = \frac{c \sin B}{\sin C} = \frac{18 \sin 71^\circ}{\sin 35^\circ} \approx \frac{18 \times 0.9455}{0.574} \approx 29.65$. 5.c) Triangle DEF: $d=22$, $e=17$, $D=48^\circ$, find angle $E$. - Use Law of Cosines or Law of Sines. - First find side $f$ opposite $F$: $f = ?$ (not given, so use Law of Cosines or Sine Law with given info). - Using Law of Sines: $\frac{d}{\sin D} = \frac{e}{\sin E}$. - $\sin E = \frac{e \sin D}{d} = \frac{17 \sin 48^\circ}{22} = \frac{17 \times 0.7431}{22} \approx 0.574$. - $E = \sin^{-1}(0.574) \approx 35.05^\circ$. 7. **Solve using Cosine Law:** 7.a) $b=14$, $c=17$, $A=52^\circ$, find $a$. - $a^2 = b^2 + c^2 - 2bc \cos A = 14^2 + 17^2 - 2 \times 14 \times 17 \times \cos 52^\circ$. - Calculate: $a^2 = 196 + 289 - 476 \times 0.6157 = 485 - 293.87 = 191.13$. - $a = \sqrt{191.13} \approx 13.82$. 7.b) Sides: 12, 18, 25, find all angles. - Use Law of Cosines for each angle. - Angle opposite 12: $A = \cos^{-1}\left(\frac{18^2 + 25^2 - 12^2}{2 \times 18 \times 25}\right) = \cos^{-1}\left(\frac{324 + 625 - 144}{900}\right) = \cos^{-1}(0.8944) \approx 26.57^\circ$. - Angle opposite 18: $B = \cos^{-1}\left(\frac{12^2 + 25^2 - 18^2}{2 \times 12 \times 25}\right) = \cos^{-1}\left(\frac{144 + 625 - 324}{600}\right) = \cos^{-1}(0.7417) \approx 42.14^\circ$. - Angle opposite 25: $C = 180^\circ - A - B = 180^\circ - 26.57^\circ - 42.14^\circ = 111.29^\circ$. 7.c) $L=70^\circ$, $|=19$, $m=15$, find side $n$. - Use Law of Cosines: $n^2 = 19^2 + 15^2 - 2 \times 19 \times 15 \times \cos 70^\circ$. - Calculate: $n^2 = 361 + 225 - 570 \times 0.3420 = 586 - 194.94 = 391.06$. - $n = \sqrt{391.06} \approx 19.78$. 8. **Solve using Sine or Cosine law:** 8.a) Sides: 13, 17, included angle 40°, find third side. - Use Law of Cosines: $c^2 = 13^2 + 17^2 - 2 \times 13 \times 17 \times \cos 40^\circ$. - $c^2 = 169 + 289 - 442 \times 0.7660 = 458 - 338.57 = 119.43$. - $c = \sqrt{119.43} \approx 10.93$. 8.b) $A=55^\circ$, $a=11$, $B=72^\circ$, find $b$. - Law of Sines: $b = \frac{a \sin B}{\sin A} = \frac{11 \sin 72^\circ}{\sin 55^\circ} = \frac{11 \times 0.9511}{0.8192} \approx 12.77$. 8.c) Sides: 9, 12, 15, find angle opposite 15. - Use Law of Cosines: $C = \cos^{-1}\left(\frac{9^2 + 12^2 - 15^2}{2 \times 9 \times 12}\right) = \cos^{-1}\left(\frac{81 + 144 - 225}{216}\right) = \cos^{-1}(0) = 90^\circ$. 9. **Sketch the graph of each function (one full cycle):** 9.a) $y = 2 \sin x$. - Amplitude = 2. - Period = $2\pi$. - Midline = $y=0$. - Max = 2, Min = -2. - Phase shift = 0. 9.b) $y = -3 \cos(x - 30^\circ)$. - Amplitude = 3. - Period = $2\pi$. - Midline = $y=0$. - Max = 3, Min = -3. - Phase shift = $+30^\circ$ (right shift). - Reflected over x-axis (negative amplitude). 9.c) $y = \sin(2x) + 1$. - Amplitude = 1. - Period = $\frac{2\pi}{2} = \pi$. - Midline = $y=1$. - Max = 2, Min = 0. - Phase shift = 0. 10. **Identify key properties of $y = -4 \sin(3(x + 20^\circ)) - 2$:** a) Amplitude = 4. b) Period = $\frac{360^\circ}{3} = 120^\circ$. c) Phase shift = $-20^\circ$ (left shift). d) Vertical shift = -2. e) Max = $-2 + 4 = 2$, Min = $-2 - 4 = -6$. 11. **Describe transformations from parent function:** a) $y = 2 \sin(x - 90^\circ)$: amplitude doubled, phase shift right 90°. b) $y = -\cos x + 3$: reflected over x-axis, vertical shift up 3. c) $y = 0.5 \sin(4x) - 1$: amplitude halved, period quartered ($90^\circ$), vertical shift down 1. 13. **Height of Ferris Wheel:** a) Graph the relation: sinusoidal with amplitude 20, midline at 25, period 90 seconds. b) Equation: $h(t) = 20 \sin\left(\frac{2\pi}{90} t - \frac{\pi}{2}\right) + 25$. c) Height after 10 seconds: - $h(10) = 20 \sin\left(\frac{2\pi}{90} \times 10 - \frac{\pi}{2}\right) + 25 = 20 \sin\left(\frac{2\pi}{9} - \frac{\pi}{2}\right) + 25$. - Calculate inside sine: approx $-0.94$. - Height $\approx 20 \times (-0.94) + 25 = 6.2$ meters. d) Maximum height occurs when sine argument = $\frac{\pi}{2}$, solve for $t$: - $\frac{2\pi}{90} t - \frac{\pi}{2} = \frac{\pi}{2} \Rightarrow t = 45$ seconds. 14. **Surveyor triangle:** a) Find third side using Law of Cosines: - $c^2 = 120^2 + 95^2 - 2 \times 120 \times 95 \times \cos 58^\circ$. - $c^2 = 14400 + 9025 - 22800 \times 0.5299 = 23425 - 12074 = 11351$. - $c = \sqrt{11351} \approx 106.5$ m. b) Find remaining angles using Law of Cosines or Sines. 15. **Boat distance problem:** - Use Law of Cosines with vectors or cosine rule. - Angle between paths = $35^\circ + 12^\circ = 47^\circ$. - Distance $d = \sqrt{40^2 + 52^2 - 2 \times 40 \times 52 \times \cos 47^\circ} \approx 33.4$ km. 16. **Sound wave $y = 3 \sin(5x - 40^\circ) + 1$:** a) Amplitude = 3. b) Period = $\frac{360^\circ}{5} = 72^\circ$. c) Phase shift = $\frac{40^\circ}{5} = 8^\circ$ right. d) Midline = 1. e) Max = 4, Min = -2.