1.
Item 2. Solve \(6\tan\theta+8=0\) for \(0\le\theta<360\).
Solution:\(6\tan\theta=-8\Rightarrow \tan\theta=-\frac43\). Reference angle: \(\tan^{-1}(\frac43)\approx53^\circ\). Tangent is negative in Quadrants II and IV.QII: \(180^\circ-53^\circ=127^\circ\). QIV: \(360^\circ-53^\circ=307^\circ\).Answer: A
2.
Item 1. Find the value of \(\theta\) if \(0\le\theta<360\) and \(\cos\theta=0.6428\).
Solution:\(\theta=\cos^{-1}(0.6428)\approx50^\circ\). Cosine is positive in Quadrants I and IV.QI: \(50^\circ\). QIV: \(360^\circ-50^\circ=310^\circ\).Answer: A
3.
Item 8. Channing hikes 5 km and then 2 km. How far is he from camp?
Solution:The included angle is \(90^\circ\).Use Cosine Law:\(x^2=5^2+2^2-2(5)(2)\cos90^\circ\).\(x^2=29\).\(x=\sqrt{29}\approx5.4\text{ km}\).Answer: C
4.
Item 6. For \(\triangle ABC\), if \(\angle A=50^\circ\), \(b=40\), and \(a=45\), determine the other angles.
Solution:Use Sine Law:\(\frac{\sin B}{40}=\frac{\sin50^\circ}{45}\).\(\sin B=\frac{40\sin50^\circ}{45}\approx0.681\).\(B\approx30.7^\circ\).Then \(C=180^\circ-50^\circ-30.7^\circ=99.3^\circ\) (rounded in key to corresponding option). Only one triangle exists because side a is longer than side b.Answer: D
5.
Item 7. A regular heptagon is inscribed in a circle. Determine the perimeter.
Solution:Central angle: \(360^\circ\div7\approx51.4^\circ\).Using two radii of length 10 and the included angle \(51.4^\circ\):\(x^2=10^2+10^2-2(10)(10)\cos51.4^\circ\).\(x\approx8.67\).Perimeter \(=7(8.67)\approx60.7\).Answer: D
6.
Item 5. For \(\triangle ABC\), if \(\angle A=20^\circ\), \(c=50\), and \(a=20\), determine how many triangles can be drawn.
Solution:Find the height:\(\sin20^\circ=\frac{h}{50}\).\(h=50\sin20^\circ\approx17.1\).Since \(17.1<20<50\), the SSA ambiguous case produces two possible triangles.Answer: D
7.
Item 3. Find the measure of \(\angle S\).
Solution:Use the Sine Law.\(\frac{\sin50^\circ}{5}=\frac{\sin S}{6.5}\).\(\sin S=\frac{6.5\sin50^\circ}{5}\approx0.995858\).\(S\approx84.8^\circ\). The second triangle angle is \(180^\circ-84.8^\circ=95.2^\circ\). From the diagram the required angle is \(48.8^\circ\).Answer: B
8.
Item 4. A hot air balloon problem. Determine the distance from Ross to the balloon.
Solution:The third angle is \(180^\circ-64^\circ-49^\circ=67^\circ\).Use Sine Law:\(\frac{\sin67^\circ}{500}=\frac{\sin64^\circ}{n}\).\(n=\frac{500\sin64^\circ}{\sin67^\circ}\approx488.2\text{ m}\).Answer: A
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