Identities in a triangle and proving claims

In a triangle $\displaystyle {\alpha+\beta+\gamma}={180^\circ}$, so $\displaystyle {\alpha}={180^\circ-(\beta+\gamma)}$. By the identities of angles adding up to $\displaystyle 180^\circ$:

• $\displaystyle {\sin\alpha}={\sin(\beta+\gamma)}$.
• $\displaystyle {\cos\alpha}={-\cos(\beta+\gamma)}$.

So one angle is written with the two others, and the sum formulas open it up.

The proof below finds $\displaystyle b$ by the law of sines and puts it into the area formula.
In triangle $\displaystyle ABC$ the side $\displaystyle a$ and the angles next to it, $\displaystyle \beta$ and $\displaystyle \gamma$, are given.

The proof below shows: if in triangle $\displaystyle ABC$ $\displaystyle {a}={2b\cos\gamma}$, then the triangle is isosceles, $\displaystyle {\beta}={\gamma}$. It goes from sides to sines by the law of sines, and uses $\displaystyle {\sin\alpha}={\sin(\beta+\gamma)}$.

The law of sines gives the length of every chord in a circle of radius $\displaystyle R$: the chord is $\displaystyle 2R$ times the sine of an inscribed angle standing on it. This helps to compare chords with identities.

An equilateral triangle $\displaystyle ABC$ is inscribed in a circle of radius $\displaystyle R$. The point $\displaystyle D$ is on the arc $\displaystyle BC$ that does not hold $\displaystyle A$.
Let $\displaystyle {\theta}={\angle BAD}$.
Prove: $\displaystyle {AD}={BD+CD}$.

• $\displaystyle {BD}={2R\sin\theta}$ (the angle $\displaystyle BAD$ stands on $\displaystyle BD$).
• $\displaystyle {CD}={2R\sin(60^\circ-\theta)}$ (the angle $\displaystyle CAD$ is $\displaystyle 60^\circ-\theta$).
• $\displaystyle {AD}={2R\sin\angle ABD}$, and $\displaystyle {\angle ABD}={\angle ABC+\angle CBD}={60^\circ+(60^\circ-\theta)}={120^\circ-\theta}$ ($\displaystyle {\angle CBD}={\angle CAD}$: they stand on the same arc).
• What is left is the identity $\displaystyle {\sin\theta+\sin(60^\circ-\theta)}={\sin(120^\circ-\theta)}$ – that is Exercise 1.

Prove the identity, which completes the proof that $\displaystyle {AD}={BD+CD}$. The card has $\displaystyle x$ in place of $\displaystyle \theta$, and the angles in degrees.

In triangle $\displaystyle ABC$, $\displaystyle {\sin\alpha}={2\sin\beta\cos\gamma}$.
Prove that the triangle is isosceles.
A hint: $\displaystyle {\sin\alpha}={\sin(\beta+\gamma)}$. In the card – $\displaystyle y$ for $\displaystyle \beta$ and $\displaystyle z$ for $\displaystyle \gamma$: show that $\displaystyle {\sin(\beta+\gamma)-2\sin\beta\cos\gamma}={\sin(\gamma-\beta)}$.

What does $\displaystyle {\sin(\gamma-\beta)}={0}$ give, when the angles are angles of a triangle?

In triangle $\displaystyle ABC$: $\displaystyle {BC}={10}$, $\displaystyle {\beta}={50^\circ}$, $\displaystyle {\gamma}={60^\circ}$.
Find the area of the triangle.

$\displaystyle ABCD$ is a square. $\displaystyle E$ is the midpoint of the side $\displaystyle BC$, and $\displaystyle F$ the midpoint of the side $\displaystyle CD$.
Work out $\displaystyle \sin\angle EAF$ without a calculator.

Explain why $\displaystyle {\tan\angle BAE}={\frac{1}{2}}$, and why $\displaystyle {\angle EAF}={90^\circ-2\angle BAE}$.

So $\displaystyle {\angle EAF}={90^\circ-2\tan^{-1}\frac{1}{2}}$. Work it out without a calculator. A hint: $\displaystyle {\sin(90^\circ-x)}={\cos x}$, and $\displaystyle {\cos2\alpha}={\frac{1-\tan^2\alpha}{1+\tan^2\alpha}}$.

Solve again another way: the square’s side is $\displaystyle 2$. Find $\displaystyle AE$, $\displaystyle AF$ and $\displaystyle EF$, and $\displaystyle \cos\angle EAF$ by the law of cosines. What is $\displaystyle \sin\angle EAF$?