Trigonometry: Sine and Cosine Rules

From $\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$, cross-multiplying gives $\sin B = \dfrac{b \sin A}{a} = \dfrac{6 \sin 30^\circ}{4} = 0.75$. (Option C inverted the ratio.)

First $C = 180^\circ - 50^\circ - 60^\circ = 70^\circ$. Then $\dfrac{a}{\sin 50^\circ} = \dfrac{20}{\sin 70^\circ}$, so $a = \dfrac{20\sin 50^\circ}{\sin 70^\circ} \approx \dfrac{20(0.766)}{0.940} \approx 16.3$. (You must find the third angle before applying the sine rule.)

$\sin B \approx 0.804$ gives an acute $B \approx 53.5^\circ$ or an obtuse $B \approx 126.5^\circ$. The obtuse case still works because $40^\circ + 126.5^\circ = 166.5^\circ < 180^\circ$, leaving a valid $C$. Both are valid (the ambiguous SSA case), so 2 triangles. (Since $b > a$, the larger angle $B$ can be obtuse.)

$\dfrac{8}{0.6} = \dfrac{10}{\sin Q}$, so $8 \sin Q = 10 \times 0.6 = 6$ and $\sin Q = \dfrac{6}{8} = 0.75$.

$\dfrac{b}{\sin 45^\circ} = \dfrac{6}{\sin 30^\circ}$, so $b = 6 \times \dfrac{\sin 45^\circ}{\sin 30^\circ} = 6 \times \dfrac{\sqrt{2}/2}{1/2} = 6\sqrt{2} \approx 8.49$.

By the cosine rule $\cos C = \dfrac{a^2 + b^2 - c^2}{2ab}$, the biggest opposite side $c$ makes $\cos C$ smallest, hence $C$ largest. So the largest angle always faces the longest side; an obtuse angle (negative cosine) can only sit opposite the longest side.

Rearranging $c^2 = a^2 + b^2 - 2ab\cos C$ gives $\cos C = \dfrac{a^2 + b^2 - c^2}{2ab}$ (the side opposite the angle carries the minus sign).

$c^2 = 3^2 + 4^2 - 2(3)(4)\cos 60^\circ = 25 - 12 = 13$, so $c = \sqrt{13} \approx 3.61$. ($5$ assumes $C = 90^\circ$ (Pythagoras); $13$ forgot the square root.)

$\cos A = \dfrac{b^2 + c^2 - a^2}{2bc} = \dfrac{49 + 64 - 25}{2(7)(8)} = \dfrac{88}{112} = \dfrac{11}{14} \approx 0.786$.

SAS has no opposite pair, so the sine rule cannot start; use the cosine rule $c^2 = a^2 + b^2 - 2ab\cos C$. (Pythagoras needs a right angle.)

$\text{Area} = \dfrac{1}{2}(10)(12)\sin C = 60\sin C$. This is greatest when $\sin C = 1$, i.e. $C = 90^\circ$, giving a maximum area of $60$ cm² (the right-angled case). (120 cm² dropped the $\tfrac{1}{2}$.)

A parallelogram's area is $ab\sin\theta$ (a diagonal splits it into two equal triangles, so there is no $\tfrac{1}{2}$): $6 \times 10 \times \sin 30^\circ = 60 \times \tfrac{1}{2} = 30$ cm². ($15$ cm² wrongly kept the $\tfrac{1}{2}$ from the triangle formula.)

$\text{Area} = \dfrac{1}{2}(10)(14)\sin 60^\circ = 70 \times \dfrac{\sqrt{3}}{2} = 35\sqrt{3} \approx 60.6$.

$20 = \dfrac{1}{2}(8)(10)\sin C = 40\sin C$, so $\sin C = \dfrac{1}{2}$ and the acute angle is $C = 30^\circ$. (There is also an obtuse solution $150^\circ$.)

$24 = \dfrac{1}{2}(8)(b)\sin 30^\circ = 2b$, so $b = 12$.

$a = 8 \times \dfrac{\sin 30^\circ}{\sin 60^\circ} = 8 \times \dfrac{1/2}{\sqrt{3}/2} = \dfrac{8}{\sqrt{3}} = \dfrac{8\sqrt{3}}{3} \approx 4.62$ (rationalised).

$\sin 150^\circ = \sin(180^\circ - 30^\circ) = \sin 30^\circ = \dfrac{1}{2}$ (sine is positive in the second quadrant).

$c^2 = 3^2 + 5^2 - 2(3)(5)\cos 120^\circ = 34 - 30(-\tfrac{1}{2}) = 34 + 15 = 49$, so $c = 7$. ($\sqrt{19}$ used $\cos 120^\circ = +\tfrac{1}{2}$ with the wrong sign; $\sqrt{34}$ dropped the cosine term.)

$d^2 = 8^2 + 5^2 - 2(8)(5)\cos 70^\circ = 89 - 80(0.342) \approx 61.6$, so $d \approx 7.9$ km. ($13.0$ just added the distances; $9.4$ used Pythagoras, ignoring the $70^\circ$ angle.)

The largest angle is opposite the longest side (11 m): $\cos C = \dfrac{6^2 + 8^2 - 11^2}{2(6)(8)} = \dfrac{-21}{96} \approx -0.219$. A negative cosine means $C$ is obtuse. (Compare $6^2 + 8^2 = 100$ with $11^2 = 121$: since $121 > 100$, the angle exceeds $90^\circ$.)

The chord and the two radii form a triangle, so by the cosine rule $d^2 = 8^2 + 8^2 - 2(8)(8)\cos 60^\circ = 128 - 128(\tfrac{1}{2}) = 64$, giving $d = 8$ cm. (A $60^\circ$ angle between equal radii makes the triangle equilateral, so the chord equals the radius.)

By the cosine rule $7^2 = x^2 + (x + 3)^2 - 2x(x + 3)\cos 60^\circ$. Since $\cos 60^\circ = \tfrac{1}{2}$, this becomes $49 = x^2 + (x^2 + 6x + 9) - (x^2 + 3x) = x^2 + 3x + 9$. So $x^2 + 3x - 40 = 0$ and $(x + 8)(x - 5) = 0$, giving $x = 5$ (a length is positive).

$\cos C = \dfrac{5^2 + 6^2 - 7^2}{2(5)(6)} = \dfrac{12}{60} = 0.2$, so $\sin C = \sqrt{1 - 0.2^2} = \dfrac{2\sqrt{6}}{5}$. Area $= \dfrac{1}{2}(5)(6)\sin C = 6\sqrt{6} \approx 14.7$. ($\sin C$ comes from the identity $\sin^2 C + \cos^2 C = 1$ once the cosine rule has given $\cos C$.)

Find the area, then multiply by the rate. Using the cosine rule, the angle $C$ between the sides $13$ and $14$ has $\cos C = \dfrac{13^2 + 14^2 - 15^2}{2(13)(14)} = \dfrac{140}{364} = \dfrac{5}{13}$, so $\sin C = \dfrac{12}{13}$ and the area is $\dfrac{1}{2}(13)(14)\cdot\dfrac{12}{13} = 84$ m². The cost is $84 \times \$8 = \$672$. ($\$84$ used the area as the cost, forgetting the $\$8$ rate.)

$c^2 = 36 + 64 - 2(6)(8)\cos 60^\circ = 100 - 48 = 52$, so $c = \sqrt{52} \approx 7.21$. Then $\sin A = \dfrac{6\sin 60^\circ}{7.21} \approx 0.720$, so $A \approx 46^\circ$.