1. **Problem 1: Solve triangle ABC with given $A=38^\circ40'$ , $a=9.72$ m, and $b=11.8$ m.** 2. Use the Law of Sines: $$\frac{a}{\sin A} = \frac{b}{\sin B}$$ 3. Calculate $\sin B$: $$\sin B = \frac{b \sin A}{a} = \frac{11.8 \times \sin 38^\circ40'}{9.72}$$ 4. Convert $38^\circ40'$ to decimal degrees: $$38 + \frac{40}{60} = 38.6667^\circ$$ 5. Calculate $\sin 38.6667^\circ \approx 0.6252$. 6. Compute $\sin B$: $$\sin B = \frac{11.8 \times 0.6252}{9.72} \approx 0.7587$$ 7. Find angle $B$: $$B = \sin^{-1}(0.7587) \approx 49^\circ 10'$$ 8. Find angle $C$: $$C = 180^\circ - A - B = 180^\circ - 38^\circ40' - 49^\circ10' = 92^\circ 10'$$ 9. Use Law of Sines to find side $c$: $$c = \frac{a \sin C}{\sin A} = \frac{9.72 \times \sin 92^\circ 10'}{\sin 38^\circ 40'}$$ 10. Calculate $\sin 92^\circ 10' \approx 0.9995$. 11. Compute $c$: $$c = \frac{9.72 \times 0.9995}{0.6252} \approx 15.54 \text{ m}$$ --- **Final answers for triangle 1:** - $A = 38^\circ 40'$ - $B = 49^\circ 10'$ - $C = 92^\circ 10'$ - $a = 9.72$ m - $b = 11.8$ m - $c \approx 15.54$ m --- Since the user asked to solve each triangle ABC that exists, but the second triangle is given only by sides $a=42.9$ m, $b=37.6$ m, $c=62.7$ m, we can solve it using Law of Cosines. 1. **Problem 2: Solve triangle with sides $a=42.9$ m, $b=37.6$ m, $c=62.7$ m.** 2. Use Law of Cosines to find angle $A$: $$\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{37.6^2 + 62.7^2 - 42.9^2}{2 \times 37.6 \times 62.7}$$ 3. Calculate squares: $$37.6^2 = 1413.76, \quad 62.7^2 = 3930.09, \quad 42.9^2 = 1840.41$$ 4. Compute numerator: $$1413.76 + 3930.09 - 1840.41 = 3503.44$$ 5. Compute denominator: $$2 \times 37.6 \times 62.7 = 4713.12$$ 6. Calculate $\cos A$: $$\cos A = \frac{3503.44}{4713.12} \approx 0.7433$$ 7. Find angle $A$: $$A = \cos^{-1}(0.7433) \approx 42^\circ 1'$$ 8. Use Law of Cosines to find angle $B$: $$\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{42.9^2 + 62.7^2 - 37.6^2}{2 \times 42.9 \times 62.7}$$ 9. Calculate numerator: $$1840.41 + 3930.09 - 1413.76 = 4356.74$$ 10. Calculate denominator: $$2 \times 42.9 \times 62.7 = 5379.66$$ 11. Calculate $\cos B$: $$\cos B = \frac{4356.74}{5379.66} \approx 0.8099$$ 12. Find angle $B$: $$B = \cos^{-1}(0.8099) \approx 36^\circ 0'$$ 13. Find angle $C$: $$C = 180^\circ - A - B = 180^\circ - 42^\circ 1' - 36^\circ 0' = 101^\circ 59'$$ --- **Final answers for triangle 2:** - $A \approx 42^\circ 1'$ - $B \approx 36^\circ 0'$ - $C \approx 101^\circ 59'$ - $a = 42.9$ m - $b = 37.6$ m - $c = 62.7$ m --- **Problem 3: Solve equation $3 \cot^2 \theta - 3 \cot \theta = 1$ for $\theta \in [0,360^\circ)$.** 1. Let $x = \cot \theta$. Rewrite equation: $$3x^2 - 3x - 1 = 0$$ 2. Use quadratic formula: $$x = \frac{3 \pm \sqrt{(-3)^2 - 4 \times 3 \times (-1)}}{2 \times 3} = \frac{3 \pm \sqrt{9 + 12}}{6} = \frac{3 \pm \sqrt{21}}{6}$$ 3. Calculate roots: - $$x_1 = \frac{3 + 4.583}{6} = 1.264$$ - $$x_2 = \frac{3 - 4.583}{6} = -0.264$$ 4. Recall $x = \cot \theta = \frac{\cos \theta}{\sin \theta}$. 5. Find $\theta$ for $\cot \theta = 1.264$: $$\theta = \cot^{-1}(1.264) = \tan^{-1}(\frac{1}{1.264}) \approx \tan^{-1}(0.791) = 38.4^\circ$$ 6. Since cotangent is positive in Quadrants I and III, solutions are: - $$\theta_1 = 38.4^\circ$$ - $$\theta_2 = 180^\circ + 38.4^\circ = 218.4^\circ$$ 7. Find $\theta$ for $\cot \theta = -0.264$: $$\theta = \cot^{-1}(-0.264) = \tan^{-1}(\frac{1}{-0.264}) = \tan^{-1}(-3.79)$$ 8. Calculate principal value: $$\tan^{-1}(-3.79) \approx -75.1^\circ$$ 9. Cotangent is negative in Quadrants II and IV, so add 180° or 360° accordingly: - $$\theta_3 = 180^\circ - 75.1^\circ = 104.9^\circ$$ - $$\theta_4 = 360^\circ - 75.1^\circ = 284.9^\circ$$ --- **Final solutions for Problem 3:** $$\theta \approx 38^\circ 24', 104^\circ 54', 218^\circ 24', 284^\circ 54'$$ --- **Problem 4: Solve equation $2 \cos 2x = \sqrt{3}$ for $x \in [0,2\pi)$.** 1. Isolate cosine: $$\cos 2x = \frac{\sqrt{3}}{2}$$ 2. Recall $\cos \alpha = \frac{\sqrt{3}}{2}$ at $\alpha = \frac{\pi}{6}, \frac{11\pi}{6}$. 3. Set $2x = \frac{\pi}{6}$ or $2x = \frac{11\pi}{6}$. 4. Solve for $x$: $$x = \frac{\pi}{12}, \frac{11\pi}{12}$$ 5. Since cosine is periodic with period $2\pi$, for $2x$ period is $2\pi$, so $x$ period is $\pi$. 6. Add $\pi$ to each solution to find all in $[0,2\pi)$: - $$x = \frac{\pi}{12}, \frac{11\pi}{12}, \frac{\pi}{12} + \pi = \frac{13\pi}{12}, \frac{11\pi}{12} + \pi = \frac{23\pi}{12}$$ --- **Final solutions for Problem 4:** $$x = \frac{\pi}{12}, \frac{11\pi}{12}, \frac{13\pi}{12}, \frac{23\pi}{12}$$