Question 1: Area of a triangle (2 points)
\(\text{Area}=\frac{1}{2}(14)(20)\sin 38^{\circ}=140\sin 38^{\circ}\approx 86.2\).
\(\approx 86.2\text{ ft}^{2}\)
Scoring: 1 point for the correct formula setup, 1 point for the correct area.
Question 2: Law of Sines, find a side (2 points)
\(\frac{b}{\sin 67^{\circ}}=\frac{18}{\sin 42^{\circ}}\), so \(b=\frac{18\sin 67^{\circ}}{\sin 42^{\circ}}\approx 24.8\).
\(b\approx 24.8\)
Scoring: 1 point for a correct proportion, 1 point for the correct side.
Question 3: Law of Sines, find angles (2 points)
\(\sin B=\frac{16\sin 75^{\circ}}{22}\approx 0.7025\), so \(B\approx 44.6^{\circ}\) (side \(b\) is shorter than side \(a\), so \(B\) is acute).
\(C=180^{\circ}-75^{\circ}-44.6^{\circ}\approx 60.4^{\circ}\).
\(B\approx 44.6^{\circ},\ C\approx 60.4^{\circ}\)
Scoring: 1 point for each angle.
Question 4: Law of Cosines, find a side (2 points)
\(c^{2}=13^{2}+17^{2}-2(13)(17)\cos 105^{\circ}\approx 458+114.4\approx 572.4\).
\(c\approx 23.9\).
\(c\approx 23.9\)
Scoring: 1 point for the correct setup (a positive result from the obtuse cosine), 1 point for \(c\).
Question 5: Largest angle of a triangle (2 points)
The largest angle is opposite 17.
\(\cos C=\frac{9^{2}+12^{2}-17^{2}}{2(9)(12)}=\frac{-64}{216}\approx -0.2963\).
\(C=\cos^{-1}(-0.2963)\approx 107.2^{\circ}\).
\(\approx 107.2^{\circ}\)
Scoring: 1 point for the Law of Cosines setup, 1 point for the angle.
Question 6: Choosing a law (2 points)
- Law of Cosines: the information is SAS, and the Law of Cosines gives the side opposite the included angle.
- Law of Sines: the known angle and its opposite side form a ratio that matches the unknown side and its opposite angle.
Scoring: 1 point for each correct law with a reason that names the given information.
Question 7: Proof of the area formula (3 points)
| Step | Statements | Reasons |
|---|---|---|
| 1 | Draw altitude \(\overline{AD}\) to \(\overline{BC}\), with \(AD=h\) | Through a point not on a line, there is exactly one perpendicular to the line |
| 2 | \(\text{Area}=\frac{1}{2}ah\) | Area of a triangle is one half base times height |
| 3 | \(\sin C=\frac{h}{b}\) | Definition of sine in right triangle ADC |
| 4 | \(h=b\sin C\) | Multiplication property of equality (multiply both sides by \(b\)) |
| 5 | \(\text{Area}=\frac{1}{2}ab\sin C\) | Substitution (replace \(h\) in step 2 with \(b\sin C\)) |
Scoring: 1 point for reasons (a) and (b), 1 point for reasons (c) and (d), 1 point for reason (e). Accept equivalent wording.
Question 8: Two ships (3 points)
The courses are two sides and the included angle, so use the Law of Cosines.
\(d^{2}=28^{2}+40^{2}-2(28)(40)\cos 64^{\circ}=2384-982.0\approx 1402.0\).
\(d\approx 37.4\).
\(\approx 37.4\text{ mi}\)
Scoring: 1 point for choosing the Law of Cosines with correct substitution, 1 point for the correct value of \(d^{2}\), 1 point for \(d\) with units.
Question 9: Area of a lot (2 points)
\(\text{Area}=\frac{1}{2}(85)(110)\sin 52^{\circ}=4675\sin 52^{\circ}\approx 3684.0\).
About 3684 square feet
Scoring: 1 point for the setup, 1 point for the rounded area.
20 points
See every chapter on the Geometry math page.