Sine and cosine rules and area of a triangleAQA A-Level Maths: Revision notes

For any angle $\theta$, take a point $(x,y)$ on a circle of radius $r$ at angle $\theta$ measured anticlockwise from the positive $x$-axis. Then $\sin\theta=\frac yr$, $\cos\theta=\frac xr$, $\tan\theta=\frac yx$. So $\sin$ is positive for $0^\circ<\theta<180^\circ$, $\cos$ is positive for acute angles and negative for obtuse angles, and $\tan\theta=\frac{\sin\theta}{\cos\theta}$. Key results: $\sin(180^\circ-\theta)=\sin\theta$ and $\cos(180^\circ-\theta)=-\cos\theta$. So $\sin150^\circ=\frac12$ and $\cos120^\circ=-\frac12$. Exact values: $\sin30^\circ=\cos60^\circ=\frac12$, $\sin60^\circ=\cos30^\circ=\frac{\sqrt3}{2}$, $\sin45^\circ=\cos45^\circ=\frac{\sqrt2}{2}$, $\tan45^\circ=1$.

For a triangle with sides $a,b,c$ opposite angles $A,B,C$: $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}.$ Use it when you know a side and the angle opposite it (a complete pair) plus one more side or angle. To find an angle use $\frac{\sin A}{a}=\frac{\sin B}{b}$. Example: in a triangle with angles $52^\circ$ and $71^\circ$ the third angle is $57^\circ$, and with that angle opposite a side of $120$ m and $AT$ opposite $71^\circ$, $\frac{AT}{\sin71^\circ}=\frac{120}{\sin57^\circ}$ gives $AT=135$ m. Take care: when you use the sine rule to find an angle, $\sin\theta=\sin(180^\circ-\theta)$, so there may be an acute and an obtuse solution. Decide which fits using the angle sum and the fact that the largest angle is opposite the longest side.

Use the cosine rule with two sides and the included angle (to find the third side), or with three sides (to find an angle): $a^2=b^2+c^2-2bc\cos A,\qquad\cos A=\frac{b^2+c^2-a^2}{2bc}.$ If $\cos A$ is negative the angle is obtuse. Example: $AB=9$, $BC=14$, $B=40^\circ$: $AC^2=81+196-252\cos40^\circ=83.96$, so $AC=9.16$ cm. For $LM=6$, $LN=10$, $MN=14$: $\cos L=\frac{36+100-196}{120}=-\frac12$, so $L=120^\circ$. The cosine rule gives a single angle, so there is no ambiguity.

The area of a triangle with two sides $a$ and $b$ and the included angle $C$ is $\text{Area}=\frac12ab\sin C.$ The angle must be between the two sides used. Example: $\frac12(9)(14)\sin40^\circ=40.5$ cm$^2$. The area can also give a perpendicular height: from $\frac12\times\text{base}\times h=\text{area}$, or $h=AT\sin52^\circ$. For an obtuse angle $\sin\theta=\sin(180^\circ-\theta)$, so $\frac12(6)(10)\sin120^\circ=15\sqrt3$.

Choose by what you know: two sides and the included angle, or three sides: cosine rule. A side and its opposite angle: sine rule. Right-angled: SOH CAH TOA. Bearings are measured clockwise from north in three figures. The bearing of $A$ from $B$ is the bearing of $B$ from $A$ plus or minus $180^\circ$. Example: $H$ to $B$ is $8$ km on $040^\circ$, then $B$ to $L$ is $11$ km on $125^\circ$: the angle $H\hat{B}L=(040^\circ+180^\circ)-125^\circ=95^\circ$, so $HL^2=8^2+11^2-2(8)(11)\cos95^\circ=200.3$ and $HL=14.2$ km. Keep full calculator values until the final answer and round at the end.