A triangle has two corners with angles of # pi / 2 # and # (3 pi )/ 8 #. If one side of the triangle has a length of #4 #, what is the largest possible area of the triangle?

Question

A triangle has two corners with angles of #  pi  /  2  # and # (3 pi  )/  8   #.  If one side of the triangle has a length of #4  #, what is the largest possible area of the triangle?

Avery Allen · Accepted Answer

The area of the triangle is #=19.3u^2#

The third angle of the triangle is

#=1/2pi-3/8pi=1/8pi#

To have the largest possible area, the side of length #4# is opposite the smallest angle, i.e, #1/8pi#

Let the side opposite the angle #3/8pi# be #=a#

Applying the sine rule to the triangle,

#a/sin(3/8pi)=4/sin(1/8pi)#

Therefore,

#a=4*sin(3/8pi)/sin(1/8pi)=9.66#

The area of the triangle is

#A=1/2*4*9.66=19.3#

Parker Allen · Answer

undefined The largest possible area of the triangle can be found using the formula for the area of a triangle, which is $ \frac{1}{2} \times \text{base} \times \text{height} $. Since one side of the triangle has a length of 4, it serves as the base. To maximize the area, we need to maximize the height of the triangle.

To maximize the height, we need to find the length of the altitude from the vertex opposite the base. Since we know two angles of the triangle, we can use trigonometric ratios to find the height.

The angle between the base and the altitude can be found by subtracting the given angles from $ \pi $, as the sum of angles in a triangle is $ \pi $ radians.

Let's denote the angle between the base and the altitude as $ \theta $. Therefore, $ \theta = \pi - \frac{\pi}{2} - \frac{3\pi}{8} $.

Now, we can use the sine function to find the height, $ h $, of the triangle:

$ h = 4 \times \sin(\theta) $.

Once we find the height, we can calculate the area of the triangle using the formula:

$ \text{Area} = \frac{1}{2} \times 4 \times h $.

By finding the maximum value of $ \text{Area} $, we can determine the largest possible area of the triangle.