A triangle has sides A, B, and C. The angle between sides A and B is #(pi)/4# and the angle between sides B and C is #pi/6#. If side B has a length of 2, what is the area of the triangle?

Question

A triangle has sides A, B, and C.  The angle between sides A and B is #(pi)/4# and the angle between sides B and C is #pi/6#.  If side B has a length of 2, what is the area of the triangle?

Emery Adkins · Accepted Answer

The area of the triangle is #=0.74u^2#

The angle between side #A# and #C# is

#=pi-(1/4pi+1/6pi)=pi-5/12pi=7/12pi#

Using the triangle and the sine rule

#A/sin(1/6pi)=B/sin(7/12pi)=2/sin(7/12pi)=2.07#

#A=2.07*sin(1/2pi)=1.04#

The triangle's area is

#=1/2*A*B*sin(1/4pi)=1/2*1.04*2*sin(1/4pi)=0.74u^2#

Charles Autterson · Answer

undefined To find the area of the triangle with sides $A$, $B$, and $C$ given the lengths of sides $B$ and the angles between them, we can use the law of sines to find the lengths of the other sides and then apply the formula for the area of a triangle.

Given:
- $B = 2$
- Angle between sides $A$ and $B$ is $\frac{\pi}{4}$ (45 degrees)
- Angle between sides $B$ and $C$ is $\frac{\pi}{6}$ (30 degrees)

Using the law of sines:
$$
\frac{A}{\sin(\frac{\pi}{6})} = \frac{2}{\sin(\frac{\pi}{4})}
$$
Solving for $A$:
$$
A = 2 \cdot \frac{\sin(\frac{\pi}{6})}{\sin(\frac{\pi}{4})}
$$

Similarly:
$$
\frac{C}{\sin(\frac{\pi}{6})} = \frac{2}{\sin(\frac{\pi}{6})}
$$
Solving for $C$:
$$
C = 2 \cdot \frac{\sin(\frac{\pi}{6})}{\sin(\frac{\pi}{6})}
$$

Now, we can use Heron's formula to find the area of the triangle:
$$
s = \frac{A + B + C}{2}
$$
$$
\text{Area} = \sqrt{s(s - A)(s - B)(s - C)}
$$

Substituting the values of $A$, $B$, and $C$, we get:
$$
s = \frac{A + 2 + C}{2}
$$
$$
\text{Area} = \sqrt{s(s - A)(s - 2)(s - C)}
$$

Calculate $s$ and substitute the values into the formula to find the area of the triangle.