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

1 Solving an AAS triangle ★★★

Angles sum to \(180^\circ\): \(C=180^\circ-52^\circ-71^\circ=57^\circ\).

Law of Sines: \(b=\dfrac{18\sin71^\circ}{\sin52^\circ}\approx 21.6\text{ cm}\) and \(c=\dfrac{18\sin57^\circ}{\sin52^\circ}\approx 19.2\text{ cm}\).

Answer: \(C=57^\circ\), \(b\approx21.6\) cm, \(c\approx19.2\) cm (the largest side, \(b\), faces the largest angle, \(71^\circ\)).

2 A missing angle ★★★

\(\sin B=\dfrac{7\sin48^\circ}{9}\approx0.578\), so \(B\approx35.3^\circ\) (the obtuse choice \(144.7^\circ\) is impossible because \(48^\circ+144.7^\circ>180^\circ\)).

\(C=180^\circ-48^\circ-35.3^\circ=96.7^\circ\) and \(c=\dfrac{9\sin96.7^\circ}{\sin48^\circ}\approx12.0\) in.

Answer: \(B\approx35.3^\circ\), \(C\approx96.7^\circ\), \(c\approx12.0\) in.

3 A side from SAS ★★★

Law of Cosines: \(a^2=12^2+15^2-2(12)(15)\cos60^\circ=144+225-180=189\).

So \(a=\sqrt{189}=3\sqrt{21}\approx13.75\).

4 Area from two sides and the angle ★★★

\(K=\dfrac12(10)(13)\sin30^\circ=\dfrac12(130)(0.5)=32.5\).

Answer: the sail has an area of \(32.5\text{ m}^2\) (about \(350\text{ ft}^2\)).

5 Heron’s formula ★★★

\(s=\dfrac{7+8+9}{2}=12\). Then \(K=\sqrt{12\cdot5\cdot4\cdot3}=\sqrt{720}=12\sqrt5\approx26.83\).

6 True or false? ★★★

1. True. \(\cos90^\circ=0\), so the term \(-2ab\cos C\) vanishes.
2. False. \(c^2-a^2-b^2=-2ab\cos C>0\) forces \(\cos C<0\), so \(C\) is obtuse.
3. False. Each ratio in the Law of Sines needs one known side-angle pair. With only sides known, there is none; use the Law of Cosines.

7 Which law first? ★★★

1. AAS pattern: find \(C=67^\circ\), then use the Law of Sines.
2. SAS pattern: the Law of Cosines gives the third side.
3. SSS pattern: the Law of Cosines gives an angle, starting with the largest.
4. SSA pattern: the Law of Sines, with a check for the ambiguous case (here \(h=11\sin25^\circ\approx4.65<8<11\), so two triangles).

8 No triangle ★★★

\(\sin B=\dfrac{10\sin40^\circ}{6}\approx1.071>1\). No angle has a sine larger than 1, so no such triangle exists.

Equivalent check: the altitude from \(C\) is \(h=10\sin40^\circ\approx6.43>6\), so side \(a\) is too short to reach the base.

9 Two triangles ★★★

\(h=11\sin35^\circ\approx6.31<8<11\): two triangles. \(\sin B=\dfrac{11\sin35^\circ}{8}\approx0.789\), so \(B_1\approx52.1^\circ\), \(B_2\approx127.9^\circ\). Both satisfy \(A+B<180^\circ\).

Triangle 1: \(C_1\approx92.9^\circ\), \(c_1=\dfrac{8\sin92.9^\circ}{\sin35^\circ}\approx13.9\).

Triangle 2: \(C_2\approx17.1^\circ\), \(c_2=\dfrac{8\sin17.1^\circ}{\sin35^\circ}\approx4.1\).

10 A right-triangle borderline ★★★

\(\sin B=\dfrac{10\sin30^\circ}{5}=1\), so \(B=90^\circ\): exactly one triangle, a right triangle (\(a=h\)).

\(C=60^\circ\) and \(c=\dfrac{5\sin60^\circ}{\sin30^\circ}=5\sqrt3\approx8.66\).

11 Across the river ★★★

\(\angle ATB=180^\circ-62^\circ-71^\circ=47^\circ\).

Law of Sines: \(AT=\dfrac{80\sin71^\circ}{\sin47^\circ}\approx103.4\text{ ft}\) and \(BT=\dfrac{80\sin62^\circ}{\sin47^\circ}\approx96.6\text{ ft}\).

Answer: the tree is about 103.4 ft (31.5 m) from A and 96.6 ft (29.4 m) from B.

12 All angles from three sides ★★★

Largest angle first: \(\cos C=\dfrac{25+49-81}{2\cdot5\cdot7}=-0.1\), so \(C\approx95.74^\circ\).

\(\cos A=\dfrac{49+81-25}{2\cdot7\cdot9}=\dfrac{105}{126}\), so \(A\approx33.56^\circ\). Then \(B=180^\circ-95.74^\circ-33.56^\circ=50.70^\circ\).

Check: \(\cos B=\dfrac{25+81-49}{2\cdot5\cdot9}=\dfrac{57}{90}\), giving \(50.70^\circ\) as well.

13 Area with two angles and a side ★★★

\(C=70^\circ\). Law of Sines: \(a=\dfrac{10\sin50^\circ}{\sin70^\circ}\approx8.152\) and \(b=\dfrac{10\sin60^\circ}{\sin70^\circ}\approx9.216\).

\(K=\dfrac12ab\sin C=\dfrac12(8.152)(9.216)\sin70^\circ\approx35.3\) square units.

14 Diagonals of a parallelogram ★★★

The diagonal facing the \(70^\circ\) angle: \(d_1^2=8^2+11^2-2(8)(11)\cos70^\circ=185-176\cos70^\circ\), so \(d_1\approx11.17\) cm.

The other diagonal faces the consecutive angle \(110^\circ\), where \(\cos110^\circ=-\cos70^\circ\): \(d_2^2=185+176\cos70^\circ\), so \(d_2\approx15.66\) cm.

15 A ship’s return trip ★★★

The back-bearing at Q is \(050^\circ+180^\circ=230^\circ\), so \(\angle PQR=230^\circ-125^\circ=105^\circ\).

\(PR^2=30^2+45^2-2(30)(45)\cos105^\circ\), so \(PR\approx60.2\) mi (about 96.9 km).

\(\cos P=\dfrac{30^2+60.2^2-45^2}{2(30)(60.2)}\approx0.692\), so \(\angle QPR\approx46.2^\circ\). The bearing from P to R is \(50^\circ+46.2^\circ=96.2^\circ\); the return bearing is \(96.2^\circ+180^\circ=\mathbf{276.2^\circ}\).

16 Spotting a fire ★★★

At A, the angle between east (\(090^\circ\)) and \(048^\circ\) is \(42^\circ\). At B, the direction to A is \(270^\circ\), and \(330^\circ-270^\circ=60^\circ\). So \(\angle AFB=180^\circ-42^\circ-60^\circ=78^\circ\).

\(AF=\dfrac{5\sin60^\circ}{\sin78^\circ}\approx4.43\text{ km}\), \(BF=\dfrac{5\sin42^\circ}{\sin78^\circ}\approx3.42\text{ km}\).

Distance north of AB: \(AF\sin42^\circ\approx2.96\text{ km}\).

17 Heron, altitude and inscribed circle ★★★

1. \(s=35\), so \(K=\sqrt{35\cdot18\cdot10\cdot7}=\sqrt{44100}=210\).
2. From \(K=\dfrac12\,\text{base}\times\text{height}\), the altitude is \(\dfrac{2K}{\text{base}}\). It is smallest on the longest side: \(h=\dfrac{420}{28}=15\). (The other altitudes are \(\dfrac{420}{17}\approx24.7\) and \(\dfrac{420}{25}=16.8\).)
3. \(r=\dfrac{210}{35}=6\).

18 A land parcel ★★★

Triangle ABC (SAS): \(AC^2=50^2+70^2-2(50)(70)\cos60^\circ=2500+4900-3500=3900\), so \(AC\approx62.45\) m. Its area is \(\dfrac12(50)(70)\sin60^\circ\approx1515.5\text{ m}^2\).

Triangle ACD (SSS, Heron): \(s=\dfrac{62.45+45+55}{2}\approx81.22\), so \(K\approx\sqrt{81.22(18.78)(36.22)(26.22)}\approx1203.6\text{ m}^2\).

Total: about \(2719\text{ m}^2\) (roughly 29,300 ft\(^2\)).

19 Recognizing a 120 degree angle ★★★

\(\cos C=\dfrac{49+64-169}{2\cdot7\cdot8}=\dfrac{-56}{112}=-\dfrac12\), so \(C=120^\circ\).

Sine formula: \(K=\dfrac12(7)(8)\sin120^\circ=28\cdot\dfrac{\sqrt3}{2}=14\sqrt3\).

Heron: \(s=14\), \(K=\sqrt{14\cdot7\cdot6\cdot1}=\sqrt{588}=14\sqrt3\). Both agree: \(K\approx24.25\).

20 How many triangles for each side? ★★★

The altitude is \(h=12\sin30^\circ=6\).

• \(a<6\): no triangle.
• \(a=6\): one triangle (right angle at B).
• \(6
• \(a\ge12\): one triangle (since \(a\ge b\), angle \(B\) cannot exceed \(A\), so the obtuse candidate is ruled out).

Sample: \(a=9\) gives \(\sin B=\dfrac{12\cdot0.5}{9}=\dfrac23\), so \(B\approx41.8^\circ\) or \(138.2^\circ\); both fit with \(30^\circ\), confirming two triangles.

21 The circumscribed circle ★★★

\(R=\dfrac{13\cdot14\cdot15}{4\cdot84}=\dfrac{2730}{336}=8.125\).

\(\sin A=\dfrac{13}{2(8.125)}=\dfrac{13}{16.25}=0.8\), so \(A\approx53.13^\circ\). Check with the Law of Cosines: \(\cos A=\dfrac{14^2+15^2-13^2}{2\cdot14\cdot15}=\dfrac{252}{420}=0.6\), and indeed \(\sin A=0.8\).