SCIDES PC11 Unit 7 Quiz 2-1 (Lesson 4-7)

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

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

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

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

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

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

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

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