A right triangle with a 30° angle

Triangle $\displaystyle ABD$ is equilateral, and each of its angles is $\displaystyle 60^{\circ}$. The altitude $\displaystyle AC$ divides it into two right triangles. An equilateral triangle is isosceles, so the altitude $\displaystyle AC$ is also an angle bisector and a median:

• an angle bisector – so $\displaystyle {\angle BAC}={60^{\circ}:2}={30^{\circ}}$;
• a median – so $\displaystyle {BC}={\frac{1}{2}BD}={\frac{1}{2}AB}$.

In the right triangle $\displaystyle ABC$, the leg $\displaystyle BC$ faces the $\displaystyle 30^{\circ}$ angle, and it is half the hypotenuse $\displaystyle AB$.

Theorem: in a right triangle with a $\displaystyle 30^{\circ}$ angle, the leg facing that angle is half the hypotenuse.

• In triangle $\displaystyle ABC$, angle $\displaystyle C$ is right and $\displaystyle {\angle A}={30^{\circ}}$.
• Then $\displaystyle {BC}={\frac{1}{2}AB}$.

Given: $\displaystyle {\angle ACB}={90^{\circ}}$ and $\displaystyle {\angle BAC}={30^{\circ}}$.
Prove: $\displaystyle {BC}={\frac{1}{2}AB}$.

The proof below completes the triangle to an equilateral one: it extends $\displaystyle BC$ past $\displaystyle C$ to the point $\displaystyle D$, so that $\displaystyle {CD}={BC}$, and joins $\displaystyle A$ to $\displaystyle D$:

• In triangle $\displaystyle ABD$, the segment $\displaystyle AC$ is an altitude and a median, so $\displaystyle {AB}={AD}$;
• all the angles of triangle $\displaystyle ABD$ are $\displaystyle 60^{\circ}$, so it is equilateral, and $\displaystyle BC$ is half its side $\displaystyle BD$.

Theorem: in a right triangle where a leg is half the hypotenuse, the angle facing that leg is $\displaystyle 30^{\circ}$.

• In triangle $\displaystyle ABC$, angle $\displaystyle C$ is right and $\displaystyle {BC}={\frac{1}{2}AB}$.
• Then $\displaystyle {\angle A}={30^{\circ}}$.

Prove the converse theorem, with the same construction: $\displaystyle D$ on the extension of $\displaystyle BC$, and $\displaystyle {CD}={BC}$.

Given: $\displaystyle {\angle ACB}={90^{\circ}}$ and $\displaystyle {AB}={2BC}$.
Prove: $\displaystyle {\angle BAC}={30^{\circ}}$.

In triangle $\displaystyle ABC$, angle $\displaystyle C$ is right, $\displaystyle {\angle A}={30^{\circ}}$, and the hypotenuse is $\displaystyle {AB}={12}$.

• The leg $\displaystyle BC$ faces the $\displaystyle 30^{\circ}$ angle, so $\displaystyle {BC}={\frac{1}{2}\cdot12}={6}$;
• the other leg, by Pythagoras’ theorem: $\displaystyle {AC}={\sqrt{12^2-6^2}}={\sqrt{108}\approx10.4}$.

In triangle $\displaystyle KLM$, angle $\displaystyle M$ is right, $\displaystyle {KL}={10}$ and $\displaystyle {LM}={5}$.

• The leg $\displaystyle LM$ is half the hypotenuse, so the angle facing it, $\displaystyle \angle K$, is $\displaystyle 30^{\circ}$;
• angle $\displaystyle L$ is $\displaystyle {180^{\circ}-90^{\circ}-30^{\circ}}={60^{\circ}}$.

Calculate, and give the reason for each step.

In triangle $\displaystyle ABC$, angle $\displaystyle C$ is right, $\displaystyle {\angle B}={60^{\circ}}$ and $\displaystyle {AB}={18}$.
Find $\displaystyle BC$.

In triangle $\displaystyle PQR$, angle $\displaystyle R$ is right, $\displaystyle {\angle P}={30^{\circ}}$ and $\displaystyle {QR}={7}$.
Find the hypotenuse $\displaystyle PQ$ and the leg $\displaystyle PR$.

In triangle $\displaystyle DEF$, angle $\displaystyle F$ is right, $\displaystyle {DE}={16}$ and $\displaystyle {EF}={8}$.
Find the angles of the triangle.

In triangle $\displaystyle ABC$, $\displaystyle {AB}={AC}={10}$, and the apex angle $\displaystyle A$ is $\displaystyle 120^{\circ}$.
The segment $\displaystyle AD$ is the altitude to the base $\displaystyle BC$.
Find $\displaystyle AD$ and $\displaystyle BC$.

In triangle $\displaystyle ABC$, angle $\displaystyle C$ is right and $\displaystyle {\angle A}={30^{\circ}}$.
The segment $\displaystyle CD$ is the altitude to the hypotenuse.
Prove: $\displaystyle {AB}={4BD}$.