Solving geometry problems with trigonometry

A problem in a polygon is solved in the triangles inside it – a diagonal, a height or a radius makes them. In each triangle choose the law by what is known:

• a pair of a side and the angle opposite it, and one more side or angle – the law of sines;
• two sides and the angle between them, or three sides – the law of cosines;
• a right angle – the ratios in a right-angled triangle.

A side shared by two triangles carries the calculation from one to the other.

In the quadrilateral $\displaystyle ABCD$: $\displaystyle {AB}={3}$, $\displaystyle {AD}={5}$, $\displaystyle {\angle A}={120^{\circ}}$.
$\displaystyle {\angle BCD}={90^{\circ}}$, and $\displaystyle {\angle DBC}={30^{\circ}}$.
The diagonal $\displaystyle BD$ cuts the quadrilateral into the triangle $\displaystyle ABD$, where two sides and the angle between them are known, and the right-angled triangle $\displaystyle BCD$.

A regular polygon with $\displaystyle n$ sides is inscribed in a circle of radius $\displaystyle R$. Each side stands on a central angle of $\displaystyle \frac{360^{\circ}}{n}$, and the inscribed angle opposite it is half of that, $\displaystyle \frac{180^{\circ}}{n}$. By the law of sines in the triangle of three neighbouring vertices:

In a hexagon: $\displaystyle {a}={2R\sin30^{\circ}}={R}$.
Work out the side of a regular octagon inscribed in a circle of radius $\displaystyle 10$. Round to two decimal places.

Sometimes a datum is missing: write it as a parameter, and express the result with it.

An isosceles triangle has legs of length $\displaystyle a$ and apex angle $\displaystyle \alpha$.

• Write the base and the area in $\displaystyle a$ and $\displaystyle \alpha$.
• At which apex angle is the area largest? Explain by the range of the sine, without differentiating.
• Check the answer: are the units of the area you found units of area?

When there are more data than needed, check that they do not contradict each other; and even three data do not always fit a triangle.

In a triangle: $\displaystyle {a}={4}$, $\displaystyle {\alpha}={30^{\circ}}$, $\displaystyle {b}={10}$.
$\displaystyle x$ is the angle opposite $\displaystyle b$. Solve the equation the law of sines gives. Is there such a triangle?

Trigonometry proves theorems too. In the drawing: $\displaystyle AD$ bisects the angle $\displaystyle A$ in triangle $\displaystyle ABC$, and $\displaystyle D$ is on $\displaystyle BC$.
The proof below shows that $\displaystyle {\frac{BD}{DC}}={\frac{AB}{AC}}$ – the angle bisector theorem – with the areas of the two triangles $\displaystyle ABD$ and $\displaystyle ADC$.

• By the area formula: $\displaystyle {S_{ABD}}={\frac{1}{2}AB\cdot AD\sin\alpha}$ and $\displaystyle {S_{ADC}}={\frac{1}{2}AC\cdot AD\sin\alpha}$, so $\displaystyle {\frac{S_{ABD}}{S_{ADC}}}={\frac{AB}{AC}}$.
• The two triangles have the same height from $\displaystyle A$ to $\displaystyle BC$, so $\displaystyle {\frac{S_{ABD}}{S_{ADC}}}={\frac{BD}{DC}}$.
• Both ratios equal the same ratio of areas: $\displaystyle {\frac{BD}{DC}}={\frac{AB}{AC}}$.

In triangle $\displaystyle ABC$ the point $\displaystyle D$ is on the side $\displaystyle BC$.
$\displaystyle {\angle BAD}={\alpha}$ and $\displaystyle {\angle DAC}={\beta}$.
Prove with the area formula: $\displaystyle {\frac{BD}{DC}}={\frac{AB\sin\alpha}{AC\sin\beta}}$.