A triangle has two corners with angles of # pi / 4 # and # (3 pi )/ 8 #. If one side of the triangle has a length of #3 #, what is the largest possible area of the triangle?

Question

A triangle has two corners with angles of #  pi  /  4  # and # (3 pi  )/  8   #.  If one side of the triangle has a length of #3  #, what is the largest possible area of the triangle?

Evelyn Arboleda · Accepted Answer

#9/4(sqrt{2}+1)#

Two of the angles given are #pi/4# and #(3pi)/8#. The third angle is then #pi - pi/4-(3pi)/8 = (3pi)/8#. Thus the triangle is an isosceles triangle. All triangles satisfying this must be similar.

Now, for the area to be as large as possible, its sides must be as large as possible. For this to happen, the one side that is specified must be the smallest one - and hence must be opposite the smallest angle - in this case #pi/4#. So, the largest triangle fitting the specifications is an isosceles triangle with base 3, and both base angles #(3pi)/8#.

The height of this triangle #= 1.5 tan((3pi)/8) = 3/2(sqrt{2}+1)#

Thus the area is #1/2 times 3 times 3/2(sqrt{2}+1) = 9/4(sqrt{2}+1)#

Note :

#tan theta = sin theta/cos theta = (2sin^2theta)/(2sin theta cos theta) = (1-cos(2theta))/sin(2theta) = csc(2theta)-cot(2theta) implies# #tan((3pi)/8) = csc((3pi)/4)-cot((3pi)/4) = sqrt(2)-(-1) = sqrt(2)+1#

Maverick Smith · Answer

undefined To find the largest possible area of the triangle, we can use the formula for the area of a triangle, which is 0.5 times the product of two sides multiplied by the sine of the included angle. Let's denote the side opposite the angle $ \frac{\pi}{4} $ as $ a $ and the side opposite the angle $ \frac{3\pi}{8} $ as $ b $. We are given that one side of the triangle has a length of 3, so we can let $ a = 3 $.

Using the Law of Sines, we can find the lengths of the other sides:

$$ \frac{a}{\sin(\frac{\pi}{4})} = \frac{b}{\sin(\frac{3\pi}{8})} $$

Solving for $ b $:

$$ b = \frac{a \cdot \sin(\frac{3\pi}{8})}{\sin(\frac{\pi}{4})} $$

$$ b = \frac{3 \cdot \sin(\frac{3\pi}{8})}{\sin(\frac{\pi}{4})} $$

$$ b \approx \frac{3 \cdot 0.3827}{0.7071} $$

$$ b \approx 1.629 $$

Now, we can use the formula for the area of a triangle:

$$ A = \frac{1}{2} \cdot a \cdot b \cdot \sin(\theta) $$

Where $ \theta $ is the angle between sides $ a $ and $ b $, which is $ \frac{\pi}{4} $.

$$ A = \frac{1}{2} \cdot 3 \cdot 1.629 \cdot \sin(\frac{\pi}{4}) $$

$$ A \approx 2.441 $$

So, the largest possible area of the triangle is approximately 2.441 square units.