A triangle has two corners with angles of # ( pi ) / 3 # and # ( pi )/ 6 #. If one side of the triangle has a length of #1 #, what is the largest possible area of the triangle?

Question

A triangle has two corners with angles of #  ( pi ) /  3  # and # ( pi  )/  6   #.  If one side of the triangle has a length of #1  #, what is the largest possible area of the triangle?

Avery Allen · Accepted Answer

If the hypotenuse is 1 then #Area_(max)=root2(3)/8#
If one of the two catheti is 1 then #Area_(max)=roots(3)/2# In any case we are dealing with a rectangle triangle. If the hypotenuse is 1, the cathetus #x# is the base whereas the cathetus #y# is the height and its area is #Area=x*y/2# under the constraint that #x^2+y^2=1#. If #y# is the cathetus opposite to the #pi/3# angle, from trigonometry we know that #y=x*tan(pi/3)=xroot2(3)#. As a result the area is #Area=x^2root2(3)/2#. But we can deduce #x^2# from the constraint and it results #3x^2+x^2=1# from which we have #x^2=1/4#. Replaced this in the area formula, finally we have #Area=root2(3)/8# If the base is 1 then #y=root2(3)# and #Area=1*y/2=root2(3)/2#

Carson Arnold · Answer

undefined To find the largest possible area of the triangle, we can use the formula for the area of a triangle, which is $ \frac{1}{2} \times \text{base} \times \text{height} $. Given that one side of the triangle has a length of 1 and the angles are $ \frac{\pi}{3} $ and $ \frac{\pi}{6} $, we can determine the height of the triangle using trigonometry.

Since the angles are $ \frac{\pi}{3} $ and $ \frac{\pi}{6} $, the side opposite the $ \frac{\pi}{3} $ angle is $ \frac{\sqrt{3}}{2} $ times the length of the side with length 1. Therefore, the height of the triangle is $ \frac{\sqrt{3}}{2} $.

Using the formula for the area of a triangle:

$$ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} $$

$$ \text{Area} = \frac{1}{2} \times 1 \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4} $$

So, the largest possible area of the triangle is $ \frac{\sqrt{3}}{4} $.