A triangle has two corners with angles of # pi / 6 # and # (3 pi )/ 8 #. If one side of the triangle has a length of #8 #, what is the largest possible area of the triangle?

Question

A triangle has two corners with angles of #  pi  /  6  # and # (3 pi  )/  8   #.  If one side of the triangle has a length of #8  #, what is the largest possible area of the triangle?

Sadie Allen · Accepted Answer

The area is #=2.75#

The angles ot the triangle are

#hatA=1/6pi#

#hatB=3/8pi#

#hatC=pi-(1/6pi+3/8pi)=pi-13/24pi=11/24pi#

The side of length is opposite the smallest angle in the triangle

So,

#a=6#

We apply the sine rule to the triangle

#b/sin hatB=a/sin hatA#

#b/sin(3/8pi)=6/sin(1/6pi)#

#b=6*sin(3/8pi)/sin(1/6pi)=11.1#

The area of the triangle is

#area =1/2absin hatC=1/2*6*11.1*sin(11/24pi)#

#=2.75#

Carson Arnold · Answer

undefined To find the largest possible area of the triangle given two angles and one side length, you can use the formula for the area of a triangle:

Area = (1/2) * base * height

In this case, the side length 8 will be the base of the triangle. To find the height, you can use trigonometry since you have the angles.

First, find the height corresponding to the angle π/6:
height_1 = 8 * tan(π/6)

Next, find the height corresponding to the angle (3π)/8:
height_2 = 8 * tan((3π)/8)

Now, calculate both heights and choose the larger one to maximize the area.

Once you have the heights, substitute them into the formula for the area and calculate the largest possible area of the triangle.