1. **Convert $\pi$ radians to grads.** Grads and radians are related by $$200 \text{ grads} = \pi \text{ radians}$$ So, $$\pi \text{ radians} = 200 \text{ grads}$$ Among the options, the closest is 250 grads (c), but mathematically it is exactly 200 grads. 2. **Convert 3200 mils to degrees.** Since 1 mil $= \frac{360^\circ}{6400}$, then $$3200 \text{ mils} = 3200 \times \frac{360}{6400} = 180^\circ$$ Answer: (a) 180° 3. **Find the third leg of the triangle with legs 7 and 4, and hypotenuse 6.84 (assuming right triangle).** Check by Pythagoras: Let the unknown leg be $x$: $$7^2 + 4^2 = 49 + 16 = 65$$ Hypotenuse squared: $$6.84^2 = 46.7856$$ Not equal, so 6.84 is likely a leg. Find missing side $x$ using: $$x^2 + 4^2 = 6.84^2 \implies x^2 = 46.7856 - 16 = 30.7856$$ $$x = \sqrt{30.7856} \approx 5.55$$ Doesn't match options, so try: $$x^2 + 7^2 = 6.84^2 \implies x^2 = 46.7856 - 49 = -2.2144$$ Negative, no. Try with 7, $x$ and hypotenuse 6.84 no fit, so likely hypotenuse is 7, another leg 6.84, find unknown leg: $$x^2 + 6.84^2 = 7^2 \implies x^2 = 49 - 46.7856 = 2.2144$$ $$x = \sqrt{2.2144} \approx 1.49$$ That is not an option. If triangle not right, but given side 36, hard to parse without clearer info. Guessing answer closest to number: (c) 6.47 4. **Given $\sin \theta = \frac{3}{5}$, find $\cos \theta$ for acute angle** Use Pythagorean identity: $$\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$$ Answer: (a) $\frac{4}{5}$ 5. **Find angle opposite the side 7 cm in right triangle with sides 7, 24, 25.** Since 25 is hypotenuse, Use sine: $$\sin \theta = \frac{7}{25}$$ Calculate angle: $$\theta = \arcsin\left(\frac{7}{25}\right) \approx \arcsin(0.28) \approx 16.3^\circ$$ Answer: (b) 16.3° 6. **Given $\sec x = 14.6401$, find $x$.** Recall $\sec x = \frac{1}{\cos x}$ so: $$\cos x = \frac{1}{14.6401} \approx 0.0683$$ Calculate angle: $$x = \arccos(0.0683) \approx 86.08^\circ$$ Answer: (a) 86.083 7. **Find the length of flagstaff on top of 10m building given elevation angles 30° to top of building, 45° to top of flagstaff from point P.** Let horizontal distance $d$ from P to building base. From angle 30°: $$\tan 30^\circ = \frac{10}{d} \Rightarrow d = \frac{10}{\tan 30^\circ} = \frac{10}{\frac{1}{\sqrt{3}}} = 10\sqrt{3} \approx 17.32 \text{ m}$$ From angle 45° to top of flagstaff of height $h$, total height $= 10 + h$: $$\tan 45^\circ = \frac{10 + h}{d} = 1 \Rightarrow 10 + h = d \Rightarrow h = d - 10 = 17.32 - 10 = 7.32 \text{ m}$$ Answer: (d) 7.32 m 8. **Building height given angles 28° from one point, 42° after moving 150 ft closer.** Let building height $h$ and initial distance $x$. From first position: $$\tan 28^\circ = \frac{h}{x} \Rightarrow h = x \tan 28^\circ$$ From second position (distance $x - 150$): $$\tan 42^\circ = \frac{h}{x - 150}$$ Equate: $$x \tan 28^\circ = (x - 150) \tan 42^\circ$$ $$x \tan 28^\circ = x \tan 42^\circ - 150 \tan 42^\circ$$ $$x(\tan 28^\circ - \tan 42^\circ) = -150 \tan 42^\circ$$ $$x = \frac{-150 \tan 42^\circ}{\tan 28^\circ - \tan 42^\circ}$$ Calculate approximate values: $$\tan 28^\circ \approx 0.5317$$ $$\tan 42^\circ \approx 0.9004$$ $$x = \frac{-150 \times 0.9004}{0.5317 - 0.9004} = \frac{-135.06}{-0.3687} \approx 366.5 \text{ ft}$$ Then, $$h = 366.5 \times 0.5317 \approx 194.8 \text{ ft}$$ Answer: (b) 194.8 ft 9. **Find angle to incline camera to see bird on 12m pole from 20m away.** $$\tan \theta = \frac{12}{20} = 0.6$$ $$\theta = \arctan(0.6) \approx 31^\circ$$ Answer: (a) 31° 10. **Find distance from wall foot given ladder length inclined at 60° reaching 6m high wall.** Ladder length $L$ such that: $$6 = L \sin 60^\circ \Rightarrow L = \frac{6}{\sin 60^\circ} = \frac{6}{\frac{\sqrt{3}}{2}} = \frac{6 \times 2}{\sqrt{3}} = \frac{12}{1.732} \approx 6.93 \, m$$ Distance from wall: $$d = L \cos 60^\circ = 6.93 \times 0.5 = 3.465 \text{ m}$$ Answer: (a) 3.464 m