A triangle has two corners with angles of # (3 pi ) / 4 # and # ( pi )/ 6 #. If one side of the triangle has a length of #9 #, what is the largest possible area of the triangle?

Question

A triangle has two corners with angles of #  (3 pi ) /  4  # and # ( pi  )/  6   #.  If one side of the triangle has a length of #9  #, what is the largest possible area of the triangle?

Theodore Abad · Accepted Answer

Largest possible area of the triangle #A_t = (1/2) b c sin A#

#color(red)(A_t = 55.3241)# Third angle #= pi - ((3pi)/4) - (pi/6) = pi / 12# #a / sin A = b / sin B = c / sin C# To get the largest area, length 9 should correspond to smallest angle C = #pi/12# #a / sin ((3pi)/4) = b / sin (pi/6) = 9 / sin (pi/12)# #a = (9 * sin ((3pi)/4)) / sin (pi/12) = 24.5885# #b = (9 * sin (pi/6)) / sin (pi/12) = 17.3867# Largest possible area of the triangle #A_t = (1/2) b c sin A# #A_t = (1/2) * 17.3867 * 9 * sin ((3pi)/4) = color(red)(55.3241)#

Mason Adams · Answer

#81/(2(sqrt3-1)) " " or " " 55.324#

First, we need to find the third angle:

#A + B + C = pi#
#(3pi)/4 + pi/6 + c = pi#
#(9pi)/12 + (2pi)/12 + c = pi#
#c = pi/12#

So our three angles are #pi/12#, #pi/6#, and #(3pi)/4#.

To get the largest possible area for the triangle, we want to make #9# the smallest side, so that the other two sides can be as big as possible.

And the smallest side is always opposite of the smallest angle. The relation between side and angle is given through the law of sines:

#a/sinA = b/sinB = c/sinC#

If we assume that the side with length #9# is opposite the angle #pi/12#, then we get:

#a/sinA = 9/sin(pi/12) = 9/((sqrt6-sqrt2)/4) = 36/(sqrt6-sqrt2)#

Now that we know this constant, we know that all of the other sides and angles must also create this same constant. So, since we have the angles for the other two sides, we can go ahead and solve for one of the other two sides, like this:

#b/sin((3pi)/4) = 36/(sqrt6-sqrt2)#
#b/(sqrt2/2) = 36/(sqrt6-sqrt2)#
#bsqrt2 = 36/(sqrt6-sqrt2)#
#b = 36/(sqrt2(sqrt6-sqrt2)) = 36/(2(sqrt3-1)) = 18/(sqrt3-1)#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Now, we know two sides, and the angle connecting them. Just for reference, this is what our triangle looks like (not to scale):

Given two sides and the angle between them, we can use this formula to find the area of the triangle:

#A_Delta = 1/2ab sinC#
#A_Delta = 1/2(9)(18/(sqrt3-1))sin(pi/6)#
#A_Delta = 1/2(9)(18/(sqrt3-1))(1/2)#
#A_Delta = 162/(4(sqrt3-1))#
#A_Delta = 81/(2(sqrt3-1)) = 55.324#

Since we made #9# the smallest side to maximize the other two lengths, this must be the largest possible area for the triangle.

Final Answer

Sadie Adams · Answer

undefined To find the largest possible area of the triangle, we need to use the formula for the area of a triangle, which is given by:

Area = (1/2) * base * height

In this case, the base will be the side of length 9, and the height will be the perpendicular distance from the vertex opposite the base to the base itself. To find the height, we can use trigonometry.

First, let's find the measure of the third angle in the triangle:

Third angle = π - (3π/4) - (π/6) = π - (9π/12) - (2π/12) = π - (11π/12) = (12π/12) - (11π/12) = π/12

Now, we'll use the angle with measure π/6 (30 degrees) to find the height. The height forms a right triangle with one side being half of the base (4.5, since 9/2 = 4.5) and the angle opposite the height being π/12.

Using trigonometry, we can use the tangent function to find the height:

tan(π/12) = height / 4.5

Solving for height:

height = 4.5 * tan(π/12)

Now, we can plug the base and height into the formula for the area of the triangle:

Area = (1/2) * 9 * (4.5 * tan(π/12))

Simplify:

Area = (1/2) * 9 * 4.5 * tan(π/12)

Area ≈ 21.22 square units