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

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

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Answer 1 · 2018-03-02T07:29:00

$34.297 {units}^{2}$ $34.297"units"^2$

Explanation:

We have a triangle that looks like:

Wikipedia

Here, we have that the angle $C = \frac{π}{2}$ $C=pi/2$ , $b = 16 units$ $b=16"units"$ and angle $A = \frac{π}{12}$ $A=pi/12$ .

Since all the angles of a triangle add up to $π$ $pi$ , angle $B = \frac{5 π}{12}$ $B=(5pi)/12$ .

According to the sine rule:

$\frac{sin A}{a} = \frac{sin B}{b}$ $sinA/a=sinB/b$

We have to solve for $a$ $a$ . Inputting:

$\frac{sin (\frac{π}{12})}{a} = \frac{sin (\frac{5 π}{12})}{16}$ $(sin(pi/12))/a=(sin((5pi)/12))/16$

$a = \frac{16 \cdot sin (\frac{π}{12})}{sin (\frac{5 π}{12})}$ $a=(16*sin(pi/12))/(sin((5pi)/12))$

$a = 4.287 units$ $a=4.287"units"$

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 $a$ $a$ as the height of the triangle and $b$ $b$ the base, and use the formula $\frac{1}{2} b h$ $1/2bh$ to solve for the area.
Find $c$ $c$ and use Heron's Formula to solve for the area.

Just for kicks, let's do both!

Use Method Number 1:

$A = \frac{1}{2} b h$ $A=1/2bh$

$A = \frac{1}{2} \cdot 16 \cdot 4.29$ $A=1/2*16*4.29$

$A = 34.297 {units}^{2}$ $A=34.297"units"^2$

Use Method Number Two:

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

Use the Pythagoras' Theorem
Use the Cosine Rule

Again, for kicks, I'm doing both.

$o \leq o \leq o \leq$ $color(white)(oleoleole)$ Sub-method Number 1:

$o \leq o \leq o \leq$ $color(white)(oleoleole)$ We have the Pythagoras' Theorem $a^{2} + b^{2} = c^{2}$ $a^2+b^2=c^2$ .

$o \leq o \leq o \leq$ $color(white)(oleoleole)$ We also know that $a = 4.29, b = 16$ $a=4.29, b=16$ . Inputting:

$o \leq o \leq o \leq$ $color(white)(oleoleole)$ $16^{2} + {4.29}^{2} = c^{2}$ $16^2+4.29^2=c^2$

$o \leq o \leq o \leq$ $color(white)(oleoleole)$ $c^{2} = 274.404$ $c^2=274.404$

$o \leq o \leq o \leq$ $color(white)(oleoleole)$ $c = 16.565 units$ $c=16.565"units"$

Now for:

$o \leq o \leq o \leq$ $color(white)(oleoleole)$ Sub-Method Number 2

$o \leq o \leq o \leq$ $color(white)(oleoleole)$ We have the cosine rule: $c^{2} = a^{2} + b^{2} - 2 a b cos C$ $c^2=a^2+b^2-2abcosC$

$o \leq o \leq o \leq$ $color(white)(oleoleole)$ We know the stuff we stated above, and $o \leq o \leq o \leq$ $color(white)(oleoleole)$ that $C = \frac{π}{2}$ $C=pi/2$

$o \leq o \leq o \leq$ $color(white)(oleoleole)$ Since $cos (\frac{π}{2}) = 0$ $cos(pi/2)=0$ , the cosine rule $o \leq o \leq o \leq$ $color(white)(oleoleole)$ $o \leq o \leq o \leq$ $color(white)(oleoleole)$ reduces to the Pythagoras' Theorem, which we $o \leq o \leq o \leq$ $color(white)(oleoleole)$ solved above, so skip it.

Now we go back to using Heron's Formula:

$A = \sqrt{s (s - a) (s - b) (s - c)}$ $A=sqrt(s(s-a)(s-b)(s-c))$ , where $s = \frac{a + b + c}{2}$ $s=(a+b+c)/2$

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

$s = \frac{4.287 + 16 + 16.565}{2}$ $s=(4.287+16+16.565)/2$

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

$A = \sqrt{18.426 (18.426 - 4.287) (18.426 - 16) (18.426 - 16.565)}$ $A=sqrt(18.426(18.426-4.287)(18.426-16)(18.426-16.565))$

$A = \sqrt{1176.216}$ $A=sqrt(1176.216)$

$A = 34.297 {units}^{2}$ $A=34.297"units"^2$

Two different methods, same answer!

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

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Explanation:

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Explanation:

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