A triangle has sides A, B, and C. The angle between sides A and B is #(2pi)/3#. If side C has a length of #2 # and the angle between sides B and C is #pi/12#, what is the length of side A?

Question

A triangle has sides A, B, and C.  The angle between sides A and B is #(2pi)/3#.  If side C has a length of #2   # and the angle between sides B and C is #pi/12#, what is the length of side A?

Zoe Austin · Accepted Answer

#0.25# #pi/12=0.261799387°=0°15'48''# #2pi/3=2.094395102=2°5'40''# Cosine rule:- #a^2=b^2+c^2-2*b*c*CosA# #b^2=a^2+c^2-2*a*c*CosB# #c^2=a^2+b^2-2*a*b*CosC# #(BC)=2*sin0261799387/sin2.094395102# #=0.00913849/0.03654595# #BC(a)= 0.25# BC=side a,AC=side b,AB = side c pi/12=.261799387° (0°15'42.18'') 2pi/3=2.094395102° (2°5'40'') 180°-(0°15'42.48'+2°5'40'')=177°38'37'' #(AC)/(sin177°38'37'')= (BA)/(sin2°5'40'')# #AC=(2*Sin177°38'37'')/(Sin2°5'40'')# #AC=0.0822/0.0365# #AC(b)=2.25# #c^2=(25)^2+(2.25)^2-2(0.25)(2.25)*0999331942# #=0.0625+50625-1.124# #AB(c)=2.0(given)

Nolan Alexander · Answer

undefined To find the length of side A in the triangle, you can use the Law of Cosines. The formula is:

$$ C^2 = A^2 + B^2 - 2AB \cdot \cos(C) $$

Given that side C has a length of 2, and the angle between sides B and C is $ \frac{\pi}{12} $, we have:

$$ 2^2 = A^2 + B^2 - 2AB \cdot \cos\left(\frac{\pi}{12}\right) $$

We also know that the angle between sides A and B is $ \frac{2\pi}{3} $, which is $ 120^\circ $ in degrees.

Now, we can solve for side A using the Law of Cosines equation. After rearranging, we get:

$$ A^2 = 2^2 + B^2 - 2 \cdot 2B \cdot \cos\left(\frac{\pi}{12}\right) $$

$$ A^2 = 4 + B^2 - 4B \cdot \cos\left(\frac{\pi}{12}\right) $$

Since the angle between sides A and B is $ \frac{2\pi}{3} $, which is $ 120^\circ $, we know $ \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} $.

Thus, we can replace $ \cos\left(\frac{\pi}{12}\right) $ with $ -\frac{1}{2} $ and solve for side A:

$$ A^2 = 4 + B^2 - 4B \cdot \left(-\frac{1}{2}\right) $$

$$ A^2 = 4 + B^2 + 2B $$

Given side C has a length of 2, using Law of Sines, we can find B:

$$ \frac{\sin(C)}{C} = \frac{\sin(B)}{B} $$

$$ \frac{\sin\left(\frac{\pi}{12}\right)}{2} = \frac{\sin(B)}{B} $$

$$ \sin(B) = \frac{B \cdot \sin\left(\frac{\pi}{12}\right)}{2} $$

$$ B = \frac{2 \cdot \sin\left(\frac{\pi}{12}\right)}{\sin(B)} $$

$$ B \approx \frac{2 \cdot 0.2588}{\sin(B)} $$

$$ B \approx \frac{0.5176}{\sin(B)} $$

$$ B \approx \frac{0.5176}{0.2588} $$

$$ B \approx 2 $$

Now substitute B = 2 into the equation for A:

$$ A^2 = 4 + 2^2 + 2(2) $$

$$ A^2 = 4 + 4 + 4 $$

$$ A^2 = 12 $$

$$ A = \sqrt{12} $$

$$ A \approx 3.4641 $$

So, the length of side A is approximately 3.4641.

Josiah Anderson · Answer

undefined You can use the Law of Cosines to find the length of side A. The formula is:

$$ A^2 = B^2 + C^2 - 2BC \cdot \cos(\theta) $$

Where:
- $ A $ is the length of side A
- $ B $ is the length of side B
- $ C $ is the length of side C
- $ \theta $ is the angle between sides B and C

Substitute the given values into the formula and solve for $ A $.