Determine the elements of a triangle knowing relationships between lengths and angles A triangle's sides' lengths are three successive numbers (meaning b=a+1, c=a+2). The smallest angle is a half of the triangle's biggest angle.
Find the area of this triangle and its angles. 
Solution: 
Sides are 4, 5 and 6.
Area: $15\sqrt{7}/4$
Angles : $41° 24' 34"$ and $82°49'8"$

Using the law of sines,
\begin{align}
   \dfrac{\sin \theta}{a} &= \dfrac{\sin 2\theta}{a+2} \\
   \dfrac{\sin \theta}{a} &= \dfrac{2 \sin(\theta) \cos(\theta)}{a+2} \\
   \cos \theta &= \dfrac{a+2}{2a} \\
\end{align}
Using the law of cosines,
\begin{align}
   a^2 &= (a+1)^2 +(a+2)^2 - 2(a+1)(a+2) \cos \theta \\
   \cos \theta &= \dfrac{(a+1)^2 +(a+2)^2 - a^2}{2(a+1)(a+2)} \\
   \cos \theta &= \dfrac{a^2+6a+5}{2(a+1)(a+2)} \\
   \cos \theta &= \dfrac{(a+1)(a+5)}{2(a+1)(a+2)} \\
   \cos \theta &= \dfrac{a+5}{2(a+2)} \\
\hline
   \dfrac{a+2}{2a} &= \dfrac{a+5}{2(a+2)} \\
   \dfrac{a+2}{a} &= \dfrac{a+5}{a+2} \\
   1 + \dfrac 2a &= 1 + \dfrac{3}{a+2} \\
   \dfrac 2a &= \dfrac{3}{a+2} \\
   3a &= 2a + 4 \\
   a &= 4
\end{align}
CHECKING
$a=4$
Smallest angle is $\theta$:
$\qquad \cos \theta = \dfrac{a+2}{2a} = \dfrac 34$  
$\qquad \sin \theta = \dfrac{\sqrt 7}{4}$
$\qquad \theta \approx 41.41^\circ$
The 'middle' angle is $180^\circ - 3\theta$
$\qquad \cos(180^\circ - 3\theta) = -\cos 3\theta = \dfrac {9}{16}$
$\qquad \sin(180^\circ - 3\theta) = \sin 3\theta = \dfrac {5\sqrt 7}{16}$
$\qquad 180^\circ - 3\theta \approx 55.77^\circ$
Largest angle is $2 \theta$:
$\qquad \sin 2\theta
  = 2 \; \sin \theta \; \cos \theta
  = \dfrac{3 \sqrt 7}{8}$
$\qquad \cos 2\theta 
  = 2\cos^2 \theta - 1
  = \dfrac 18$
$\qquad  2\theta \approx 82.82^\circ$
