Use the cosine rule to find a side given two sides and the included angle, or an angle given three sides

Use $c^2 = a^2 + b^2 - 2 a b \cos C$.

$c^2 = 49 + 81 - 2 \times 7 \times 9 \times \cos 65\degree$.

$\cos 65\degree \approx 0.4226$.

$c^2 = 130 - 126 \times 0.4226 = 130 - 53.25 = 76.75$.

$c = \sqrt{76.75} \approx 8.76$.

Rounded to one decimal place: $c \approx 8.8$ cm.

Markers reward the cosine rule stated, correct substitution, intermediate computation kept to at least four decimal places, and final answer at the requested precision.

The largest angle is opposite the longest side ($80$ m). Use the cosine rule rearranged:

$\cos A = \frac{b^2 + c^2 - a^2}{2 b c}$.

Let $a = 80$, $b = 60$, $c = 50$.

$\cos A = \frac{60^2 + 50^2 - 80^2}{2 \times 60 \times 50} = \frac{3600 + 2500 - 6400}{6000} = \frac{-300}{6000} = -0.05$.

$A = \cos^{-1}(-0.05) \approx 92.87\degree$, round to $93\degree$.

Markers reward identifying the largest angle as opposite the longest side, the rearranged cosine rule, and the angle rounded as requested. Half a mark if you find $A$ but do not justify which side it sits opposite.

State the rule (SAS). The angle $C$ lies between the two known sides, so use the side form:

Substitute the given values. With $a = 6$, $b = 10$ and $\cos 50^\circ \approx 0.6428$:

Take the square root.

Answer: $c \approx 7.7$ cm.

State the rule (SSS). Three sides and no angle means use the angle form, with the wanted angle's opposite side as $a = 4$ and the enclosing sides as $b = 5$ and $c = 6$:

Substitute the given values.

Invert to find the angle.

Answer: the angle opposite the $4$ cm side is about $41^\circ$.

State the rule (SAS). The angle $C$ sits between the two known sides, so use the side form of the cosine rule:

Substitute the given values. With $a = 8$, $b = 11$ and $C = 40\degree$, and using $\cos 40\degree \approx 0.7660$:

Take the square root.

Check. The included angle is acute, so $c$ should be shorter than the Pythagoras value $\sqrt{8^2 + 11^2} \approx 13.60$ cm, and it should sit between the difference of the sides ($3$ cm) and their sum ($19$ cm). Since $7.08$ cm satisfies both, the answer is reasonable.

Identify the largest angle. The largest angle lies opposite the longest side, which is the $9$ m side. Put that side as $a = 9$ and the two enclosing sides as $b = 6$ and $c = 5$.

State the rule (SSS). Three sides and no angle means use the angle form:

Substitute the given values.

Invert to find the angle.

Check. The numerator $b^2 + c^2 - a^2$ is negative, which predicts an obtuse angle, and $109\degree$ is indeed obtuse. The largest angle of the garden bed is about $109\degree$.

Label the SAS triangle. The angle $C = 95^\circ$ sits between the two known sides $a = 12$ and $b = 15$, so the side $c$ opposite $C$ comes from the side form.

Mind the obtuse angle. Because $95^\circ$ is obtuse, $\cos 95^\circ \approx -0.0872$ is negative, so the last term is added rather than subtracted:

Take the square root.

Check. The angle is just past a right angle, so $c$ should be a little longer than the Pythagoras value $\sqrt{12^2 + 15^2} \approx 19.21$ cm, and $20.01$ cm is, as expected. The third side is about $20.01$ cm.

Label the SAS triangle. Put the corner post at vertex $C$. The two fences are the sides $a = 45$ and $b = 70$, and the angle between them is the included angle $C = 115\degree$. The distance we want, $c$, is opposite that angle.

State and substitute the side form. Because $115\degree$ is obtuse, $\cos 115\degree \approx -0.4226$ is negative, so the last term is added:

Take the square root.

Check. With a wide obtuse angle the far ends should be well separated, close to the sum of the fences ($115$ m) rather than their difference ($25$ m), and $97.92$ m sits sensibly in that range. The ends are about $97.92$ m apart.

Identify the smallest angle. The smallest angle lies opposite the shortest side, which is the $13$ cm side. Put that side as $a = 13$ and the enclosing sides as $b = 14$ and $c = 15$.

State the rule (SSS).

Substitute the given values.

Invert to find the angle.

Check. The numerator is positive, so the angle is acute, and $53\degree$ is the smallest of the three angles because it faces the shortest side. The smallest angle is about $53\degree$.

Find the interior angle at the bend

The bearing turns by $150\degree - 070\degree = 80\degree$. The interior angle of the triangle at the turning point is the supplement, $180\degree - 80\degree = 100\degree$, because the new leg makes an $80\degree$ angle with the straight-ahead direction.

Set up the SAS triangle

The two legs $18$ km and $24$ km enclose the $100\degree$ angle, and the unknown $c$ is the distance from the campsite.

State and substitute the side form

Since $\cos 100\degree \approx -0.1736$, the last term is added:

Take the square root.

Check. The result lies between the difference of the legs ($6$ km) and their sum ($42$ km), as it must, so the walker finishes about $32.40$ km from the campsite.

Part (a), label the SAS triangle. The included angle $P = 78\degree$ sits between the two known sides $PQ = 95$ and $PR = 120$, so the side $QR$ opposite $P$ comes straight from the side form.

Substitute and solve for $QR$. Using $\cos 78\degree \approx 0.2079$:

Part (b), choose the area formula. We already know two sides and the angle between them, so the included-angle area formula is the quickest route:

Substitute. Using $\sin 78\degree \approx 0.9781$:

Check. The third side $QR \approx 136.69$ m is longer than each given side, which fits an angle near $78\degree$ opening the triangle wide, and the area is a little under the right-angled maximum of $\tfrac{1}{2} \times 95 \times 120 = 5700$ m$^2$, exactly as expected when the angle is just under $90\degree$.

Find the interior angle at $B$

The bearing changes by $110\degree - 035\degree = 75\degree$, so the interior angle of triangle $ABC$ at the bend is the supplement, $180\degree - 75\degree = 105\degree$.

Set up the SAS triangle

The two legs $AB = 60$ km and $BC = 90$ km enclose the $105\degree$ angle at $B$, and the unknown is the side $AC$ opposite $B$.

State and substitute the side form

Because $105\degree$ is obtuse, $\cos 105\degree \approx -0.2588$ is negative and the final term is added:

Take the square root.

Check. The distance lies between the difference of the legs ($30$ km) and their sum ($150$ km), and the obtuse turn pushes it towards the upper end, so $AC \approx 120.40$ km is sensible.

Part (a), set up the angle form. Put the side opposite the wanted angle as $a = 12$, and the two enclosing sides as $b = 7$ and $c = 8$:

Read the sign. The numerator $b^2 + c^2 - a^2 = -31$ is negative, so $\cos A$ is negative and the angle must lie between $90\degree$ and $180\degree$. The angle opposite the $12$ m side is therefore obtuse.

Part (b), invert to find the angle.

Convert the decimal part to minutes. The whole-degree part is $106\degree$, and $0.0685\degree \times 60 \approx 4'$, so

Check. The angle is obtuse as shown in part (a), and $106\degree$ is the largest angle of the triangle because it faces the longest side, which is consistent. The angle is about $106\degree 04'$.

Part (a), label the SAS triangle. The angle $S = 62^\circ$ sits between the two known sides $SM = 110$ and $ST = 140$, so the side $MT$ opposite $S$ comes from the side form.

Substitute and solve for $MT$. Using $\cos 62^\circ \approx 0.4695$:

Part (b), set up the angle form at $M$. Now all three sides are known, so use the angle form with the wanted angle's opposite side as $ST = 140$, and the enclosing sides $SM = 110$ and $MT \approx 131.30$:

Invert to find the angle.

Check. The three angles should sum to $180^\circ$. With $S = 62^\circ$ and $M \approx 70^\circ$, the angle at $T$ would be about $48^\circ$, and $62 + 70 + 48 = 180$, so the working is consistent. The angle $SMT$ is about $70^\circ$.