A triangle has sides A, B, and C. The angle between sides A and B is #(pi)/2# and the angle between sides B and C is #pi/12#. If side B has a length of 16, what is the area of the triangle?

1 Answer
Mar 2, 2018

#34.297"units"^2#

Explanation:

We have a triangle that looks like:

Wikipedia

Here, we have that the angle #C=pi/2#, #b=16"units"# and angle #A=pi/12#.

Since all the angles of a triangle add up to #pi#, angle #B=(5pi)/12#.

According to the sine rule:

#sinA/a=sinB/b#

We have to solve for #a#. Inputting:

#(sin(pi/12))/a=(sin((5pi)/12))/16#

#a=(16*sin(pi/12))/(sin((5pi)/12))#

#a=4.287"units"#

Now there exist possibilities all around. There are a multitude of ways to solve for the area. Let's look at the two main ways of action.

  • Take #a# as the height of the triangle and #b# the base, and use the formula #1/2bh# to solve for the area.
  • Find #c# and use Heron's Formula to solve for the area.

Just for kicks, let's do both!

Use Method Number 1:

#A=1/2bh#

#A=1/2*16*4.29#

#A=34.297"units"^2#

Use Method Number Two:

We must find #c# first. To do this, too, we have two methods:

Again, for kicks, I'm doing both.

#color(white)(oleoleole)#Sub-method Number 1:

#color(white)(oleoleole)#We have the Pythagoras' Theorem #a^2+b^2=c^2#.

#color(white)(oleoleole)#We also know that #a=4.29, b=16#. Inputting:

#color(white)(oleoleole)##16^2+4.29^2=c^2#

#color(white)(oleoleole)##c^2=274.404#

#color(white)(oleoleole)##c=16.565"units"#

Now for:

#color(white)(oleoleole)#Sub-Method Number 2

#color(white)(oleoleole)#We have the cosine rule: #c^2=a^2+b^2-2abcosC#

#color(white)(oleoleole)#We know the stuff we stated above, and #color(white)(oleoleole)#that #C=pi/2#

#color(white)(oleoleole)#Since #cos(pi/2)=0#, the cosine rule #color(white)(oleoleole)##color(white)(oleoleole)#reduces to the Pythagoras' Theorem, which we #color(white)(oleoleole)#solved above, so skip it.

Now we go back to using Heron's Formula:

#A=sqrt(s(s-a)(s-b)(s-c))#, where #s=(a+b+c)/2#

Here, #a=4.287, b=16, c=16.565#. So:

#s=(4.287+16+16.565)/2#

#s=18.426#. Inputting all of that into Heron's Formula:

#A=sqrt(18.426(18.426-4.287)(18.426-16)(18.426-16.565))#

#A=sqrt(1176.216)#

#A=34.297"units"^2#

Two different methods, same answer!