Area of a triangle using sineIB MYP Maths Extended: Revision notes

For a triangle with two sides $a$ and $b$ and the angle $C$ between them (the included angle): $\text{Area}=\frac12ab\sin C.$ It comes from $\frac12\times\text{base}\times\text{height}$: if $b$ is the base, the height is $a\sin C$. Worked example. $AB=10$ cm, $BC=14$ cm and $A\hat BC=30^\circ$: area $=\frac12(10)(14)\sin30^\circ=35$ cm$^2$.

You can rearrange the formula. If the area and two sides are known: $\sin C=\frac{2\times\text{Area}}{ab}.$ If the area is known, the perpendicular height from a vertex to the opposite side comes from $\frac12\times\text{base}\times h=\text{Area}$. For the triangle above, $\frac12(14)h=35$, so the shortest distance from $A$ to $BC$ is $h=5$ cm. An angle found from its sine may be acute or obtuse (for example $30^\circ$ and $150^\circ$ have the same sine). Use the situation to decide.

Often you must find a missing side or angle first.

• Two sides and the included angle: area immediately. Use the cosine rule if you need the third side.
• Two angles and a side: find the third angle, use the sine rule to find another side, then the area. Worked example. $A=40^\circ$, $B=60^\circ$ and $a=7$. Then $C=80^\circ$, $b=\frac{7\sin60^\circ}{\sin40^\circ}=9.43$, and area $=\frac12(7)(9.43)\sin80^\circ=32.5$. Keep full calculator values and round at the end.

A bearing is an angle measured clockwise from north, written with three figures (for example $040^\circ$). The back bearing from $B$ to $A$ is the bearing from $A$ to $B$ plus or minus $180^\circ$. Worked example. A ship sails 8 km on $040^\circ$ from $A$ to $B$, then 6 km on $100^\circ$ to $C$. The back bearing from $B$ to $A$ is $220^\circ$, so $A\hat BC=220^\circ-100^\circ=120^\circ$. Then area $=\frac12(8)(6)\sin120^\circ=20.8$ km$^2$ and $AC^2=64+36-96\cos120^\circ=148$, so $AC=12.2$ km. Draw a north line at every point and mark the angles.

A diagonal splits a parallelogram into two congruent triangles. The area of a parallelogram with sides $a$ and $b$ and included angle $\theta$ is $ab\sin\theta$. For sides 12 and 9 and $\theta=70^\circ$ that is $101.5$ cm$^2$. In design problems, compare each calculated value with the conditions given. For a sail with edges 10 m and 12 m, the area is $60\sin\theta$ and the third edge comes from the cosine rule. At $\theta=110^\circ$ the area is $56.4$ m$^2$ but the third edge is $18.1$ m, so a limit of 17 m is broken. Always finish with a conclusion in words.