# A triangle has two corners with angles of  pi / 12  and  pi / 12 . If one side of the triangle has a length of 5 , what is the largest possible area of the triangle?

Oct 16, 2017

Largest area possible $\ast 6.2506 \ast$

#### Explanation:

Three angles are $\frac{\pi}{12} , \frac{\pi}{12} , \frac{5 \pi}{6}$
$\frac{a}{\sin} a = \frac{b}{\sin} b = \frac{c}{\sin} c$

To obtain largest possible area,
Length 5 should be opposite to the angle with least value.
$\frac{5}{\sin} \left(\frac{\pi}{12}\right) = \frac{b}{\sin} \left(\frac{\pi}{12}\right) = \frac{c}{\sin} \left(\frac{5 \pi}{6}\right)$
Side b = 5.
Side $c = \frac{5 \cdot \sin \left(\frac{5 \pi}{6}\right)}{\sin} \left(\frac{\pi}{12}\right)$
Side $c = \frac{5 \cdot \sin \left(\frac{\pi}{6}\right)}{\sin} \left(\frac{\pi}{12}\right) = \frac{5}{2 \cdot \sin \left(\frac{\pi}{12}\right)}$
$c = 9.6593$
Area
$s = \frac{5 + 5 + 9.6593}{2} = 9.8297$
$A = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$

$= \sqrt{9.8297 \cdot 4.8297 \cdot 4.8297 \cdot 0.1704}$

Area $A = 6.2506$

Oct 16, 2017

Area = 6.25

#### Explanation:

Alternate method :
Area of $\Delta = \left(\frac{1}{2}\right) b h$
Given triangle is isosceles as two angles are $\frac{\pi}{12}$ each.
$\therefore h = 5 \cdot \sin \left(\frac{\pi}{12}\right)$
$\left(\frac{1}{2}\right) b . = 5 \cdot \cos \left(\frac{\pi}{12}\right)$
Area $= 5 \cdot 5 \cdot \sin \left(\frac{\pi}{12}\right) \cdot \cos \left(\frac{\pi}{12}\right)$
$= \left(\frac{25}{2}\right) \cdot 2 \sin \left(\frac{\pi}{12}\right) \cos \left(\frac{\pi}{12}\right)$
$= \left(\frac{25}{2}\right) \sin \left(\frac{\pi}{6}\right) = \left(\frac{25}{2}\right) \left(\frac{1}{2}\right) = \frac{25}{4} = 6.25$