The sine and cosine rules

The side you want, $AB$, is opposite the $37^\circ$ angle at $C$. Your angle-side pair is $BC=12$, opposite $A=30^\circ$.

$$\frac{AB}{\sin37^\circ}=\frac{12}{\sin30^\circ}\ \Rightarrow\ AB=\frac{12\sin37^\circ}{\sin30^\circ}\approx14.44$$

(C) is what you get if you flip the fraction and put $\sin37^\circ$ on the bottom of the wrong side.

The 8 cm side is opposite the $40^\circ$ angle, so that is your angle-side pair. The 10 cm side is opposite $a$.

$$\frac{\sin a}{10}=\frac{\sin40^\circ}{8}\ \Rightarrow\ \sin a=\frac{10\sin40^\circ}{8}\approx0.8035$$

So $a\approx53.5^\circ$. Because $\sin a<1$, find the second option: $180^\circ-53.5^\circ=126.5^\circ$. Now check each one with the $40^\circ$ angle: $40^\circ+53.5^\circ=93.5^\circ$ and $40^\circ+126.5^\circ=166.5^\circ$. Both are under $180^\circ$, so both triangles exist.

(C) includes $86.5^\circ$, which is the third angle of the first triangle. It is not a value for $a$.

You have all three sides and no angles, so there is no angle-side pair. Angle $A$ is opposite $CB=8.8$, so 8.8 is the side that gets subtracted:

$$\cos A=\frac{9.9^{2}+11.3^{2}-8.8^{2}}{2(9.9)(11.3)}\approx0.6627\ \Rightarrow\ A\approx48.5^\circ$$

(D) is $180^\circ-48.5^\circ$. You get it when a sign goes wrong in the numerator and $\cos A$ comes out negative.

Split the quadrilateral along the diagonal $BD$. Triangle $ABD$ has two sides and the angle between them, so start there with the cosine rule:

$$BD^{2}=4.1^{2}+7.6^{2}-2(4.1)(7.6)\cos117^\circ\ \Rightarrow\ BD\approx10.142\text{ cm}$$

Triangle $BCD$ now has an angle-side pair: $BD$, opposite the $62^\circ$ angle at $C$. Use the sine rule to find $\angle BDC$, which is opposite $BC=5.4$:

$$\sin\angle BDC=\frac{5.4\sin62^\circ}{10.142}\approx0.4701\ \Rightarrow\ \angle BDC\approx28.04^\circ$$

The other option, $151.96^\circ$, plus $62^\circ$ is over $180^\circ$, so there is only one triangle. That leaves $\angle CBD=180^\circ-62^\circ-28.04^\circ=89.96^\circ$.

$$CD=\frac{10.142\sin89.96^\circ}{\sin62^\circ}\approx11.49\text{ cm}$$

Perimeter: $4.1+5.4+11.49+7.6\approx28.59$ cm. Keep $BD$ unrounded in your calculator. Rounding it to 10.1 early moves the final answer.

The full walkthrough of 2020 Paper 2, Question 18 goes through it mark by mark.

The two 4 m sides have the $120^\circ$ angle between them.

$$\text{Area}=\frac{1}{2}(4)(4)\sin120^\circ=8\times\frac{\sqrt{3}}{2}=4\sqrt{3}$$

$\sin120^\circ=\sin60^\circ$, because $120^\circ=180^\circ-60^\circ$. (C) forgets the $\frac{1}{2}$.

The segment is the sector minus the triangle. With $\theta$ in radians, the sector has area $\frac{1}{2}r^{2}\theta$. The triangle has two sides of length $r$ with $\theta$ between them, so its area is $\frac{1}{2}r^{2}\sin\theta$.

$$\frac{1}{2}r^{2}\theta-\frac{1}{2}r^{2}\sin\theta=\frac{r^{2}}{2}\big(\theta-\sin\theta\big)$$

(A) converts $\theta$ from degrees to radians, but it is already in radians.

The 3 m and 6 m sides meet at $B$, with the $30^\circ$ angle between them.

$$\text{Area}=\frac{1}{2}(3)(6)\sin30^\circ=9\times\frac{1}{2}=\frac{9}{2}\text{ m}^{2}$$

Show the substitution and the exact value of $\sin30^\circ$. On a Technology-free paper, those steps are where the marks are.

(a) Angle $C$ is $120^\circ$, which is obtuse. A triangle can only have one obtuse angle, so the two equal angles must be $A$ and $B$, each $\frac{180^\circ-120^\circ}{2}=30^\circ$. Equal angles sit opposite equal sides, so $a=b=4$ cm.

(b) Sides $a$ and $b$ meet at $C$, so use the $120^\circ$ angle:

$$\text{Area}=\frac{1}{2}(4)(4)\sin120^\circ=8\times\frac{\sqrt{3}}{2}=4\sqrt{3}\text{ cm}^{2}$$

(a) Triangle $PQR$ has two sides and the angle between them, so use the cosine rule:

$$PR^{2}=900^{2}+540^{2}-2(900)(540)\cos65^\circ\ \Rightarrow\ PR\approx831.15\text{ m}$$

(b) The same two sides and angle give the area of section A:

$$\text{Area of A}=\frac{1}{2}(900)(540)\sin65^\circ\approx220\,233\text{ m}^{2}$$

(c) In triangle $PRS$ you know the fence and the $110^\circ$ angle opposite it, which is an angle-side pair. Use the sine rule to find $\angle SPR$, which is opposite $RS=500$:

$$\sin\angle SPR=\frac{500\sin110^\circ}{831.15}\approx0.5653\ \Rightarrow\ \angle SPR\approx34.42^\circ$$

The second option, $145.58^\circ$, is far too big to sit alongside $110^\circ$, so there is one triangle. Then $\angle PRS=180^\circ-110^\circ-34.42^\circ=35.58^\circ$, which is the angle between $RS$ and the fence:

$$\text{Area of }PRS=\frac{1}{2}(500)(831.15)\sin35.58^\circ\approx120\,891\text{ m}^{2}$$

Total: $220\,233+120\,891\approx341\,124$ m².

Start with a diagram, with a north line at each point.

At $L$, the bearing back to $H$ is $52^\circ+180^\circ=232^\circ$. The path to $C$ is on $210^\circ$, so the angle between them is $\angle HLC=232^\circ-210^\circ=22^\circ$. That is two sides and the angle between them, so use the cosine rule:

$$HC^{2}=8^{2}+5.2^{2}-2(8)(5.2)\cos22^\circ\ \Rightarrow\ HC\approx3.728\text{ km}$$

Now you need the angle at $C$. $\angle HCL$ is opposite the longest side, $HL=8$, so it could be obtuse, and the sine rule would hide that from you. Find $\angle LHC$ instead. It is opposite the shorter side $LC=5.2$, so it has to be acute:

$$\sin\angle LHC=\frac{5.2\sin22^\circ}{3.728}\approx0.5225\ \Rightarrow\ \angle LHC\approx31.5^\circ$$

So $\angle HCL=180^\circ-22^\circ-31.5^\circ=126.5^\circ$.

At $C$, the bearing back to $L$ is $210^\circ-180^\circ=030^\circ$. $H$ is $126.5^\circ$ anticlockwise from that line, so the bearing of $H$ from $C$ is $30^\circ-126.5^\circ+360^\circ=263.5^\circ$.

If you use the sine rule for $\angle HCL$ directly, your calculator gives $53.5^\circ$ and the bearing comes out wrong. It is the ambiguous case turning up where you least expect it.

The full walkthrough of 2022 Paper 2, Question 15 goes through it mark by mark.

Split it into flat triangles. The two vertical triangles, $ACD$ and $BCD$, are right-angled at $C$, so plain Year 10 trigonometry gives the ground distances:

$$AC=\frac{60}{\tan28^\circ}\approx112.84\text{ m}\qquad BC=\frac{60}{\tan35^\circ}\approx85.69\text{ m}$$

Now the triangle on the ground, $ABC$. You need $\angle ACB$, so use back bearings from $C$. The bearing of $A$ from $C$ is $50^\circ+180^\circ=230^\circ$, and the bearing of $B$ from $C$ is $300^\circ-180^\circ=120^\circ$. The angle between them is $\angle ACB=230^\circ-120^\circ=110^\circ$.

$$AB^{2}=112.84^{2}+85.69^{2}-2(112.84)(85.69)\cos110^\circ\ \Rightarrow\ AB\approx163.4\text{ m}$$

The full walkthrough of 2021 Paper 2, Question 16 goes through it mark by mark.