Law of Sines and Law of Cosines: math test solutions, Grade 12 – download the PDF

1 Law of Sines / 4 pts

\(C=180^\circ-38^\circ-64^\circ=78^\circ\) (1 pt).

\(b=\dfrac{11\sin64^\circ}{\sin38^\circ}\approx16.1\text{ cm}\) (1.5 pts).

\(c=\dfrac{11\sin78^\circ}{\sin38^\circ}\approx17.5\text{ cm}\) (1.5 pts).

2 The ambiguous case / 4 pts

(a) \(h=18\sin25^\circ\approx7.61<9<18\): two triangles (1 pt).

(b) \(\sin B=\dfrac{18\sin25^\circ}{9}\approx0.845\), so \(B_1\approx57.7^\circ\) and \(B_2\approx122.3^\circ\) (1 pt).

Triangle 1: \(C_1=180^\circ-25^\circ-57.7^\circ=97.3^\circ\) and \(c_1=\dfrac{9\sin97.3^\circ}{\sin25^\circ}\approx21.1\) (1 pt).

Triangle 2: \(C_2=180^\circ-25^\circ-122.3^\circ=32.7^\circ\) and \(c_2=\dfrac{9\sin32.7^\circ}{\sin25^\circ}\approx11.5\) (1 pt).

3 Three sides / 4 pts

(a) \(\cos C=\dfrac{81+144-256}{2\cdot9\cdot12}=-\dfrac{31}{216}\), so \(C\approx98.3^\circ\) (1.5 pts).

(b) \(\cos B=\dfrac{81+256-144}{2\cdot9\cdot16}\approx0.670\), so \(B\approx47.9^\circ\); then \(A=180^\circ-98.3^\circ-47.9^\circ\approx33.8^\circ\) (1 pt).

(c) \(s=18.5\), \(K=\sqrt{18.5\cdot9.5\cdot6.5\cdot2.5}\approx53.44\) (1.5 pts).

4 Two sides and the included angle / 4 pts

\(c^2=225+484-660\cos57^\circ\approx349.5\), so \(c\approx18.7\) (1.5 pts).

\(K=\dfrac12(15)(22)\sin57^\circ\approx138.4\) (1 pt).

\(\sin A=\dfrac{15\sin57^\circ}{18.7}\approx0.673\), so \(A\approx42.3^\circ\) (acute, since \(a\) is not the longest side) (1.5 pts).

5 Two ships / 4 pts

(a) \(\angle AHB=100^\circ-20^\circ=80^\circ\) (1 pt). \(AB^2=60^2+85^2-2(60)(85)\cos80^\circ\approx9053.8\), so \(AB\approx95.2\) km (1 pt).

(b) \(\cos A=\dfrac{60^2+95.15^2-85^2}{2(60)(95.15)}\approx0.475\), so \(\angle HAB\approx61.6^\circ\) (1 pt). The direction from A back to H is \(200^\circ\); ship B lies \(61.6^\circ\) counterclockwise from it, so the bearing is \(200^\circ-61.6^\circ=138.4^\circ\) (1 pt).