Question

Solve for the remaining angles and side of the two triangles that can be created. Round to the nearest hundredth:C = 35", a = 8.c = 5

Answer 1

Explanation

The sum of the three internal angles of a triangle is equal to 180°. Then for a triangle with angles A, B and C we get:

$A+B+C=180^{\circ}$

Considering the sides of the triangle are a, b and c and they are opposite to angles A, B and C respectively the law of sines states that:

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

Now let's use these properties to find the other two possible triangles. From the law of sines we get:

$\frac{8}{\sin A}=\frac{b}{\sin B}=\frac{5}{\sin35^{\circ}}$

We can build an equation for A:

$\frac{8}{\sin A}=\frac{5}{\sin35^{\circ}}$

We can invert both sides of this equation:

$\frac{\sin A}{8}=\frac{\sin 35^{\circ}}{5}$

And we multiply both sides by 8:

$\begin{gathered} \frac{\sin A}{8}\cdot8=\frac{\sin35^{\circ}}{5}\cdot8 \\ \sin A=\frac{8}{5}\sin 35^{\circ} \end{gathered}$

We can use the arcsin function to find A:

$\begin{gathered} \sin ^{-1}(\sin A)=\sin ^{-1}(\frac{8}{5}\sin 35^{\circ}) \\ A=66.60^{\circ} \end{gathered}$

However that's not the only possible value of A. A can be acute (smaller than 90°) or obtuse (greater than 90°). This is the value of A if it's acute. In order to find the value of A when it's obtuse we have to find an angle with a sine that is equal to that of A.

It's important to remember that for any angle x in the first quadrant [0°,90°) there's another angle in the second quadrant [90°,180°) that has the same sine. This angle is given by:

$180^{\circ}-x$

So the obtuse value of A is:

$180^{\circ}-66.60^{\circ}=113.40^{\circ}$

Now let's find B and b in each case.

First if A=66.60° the sum of A, B and C is:

$\begin{gathered} A+B+C=180^{\circ} \\ 66.60^{\circ}+B+35^{\circ}=180^{\circ} \\ B+101.60^{\circ}=180^{\circ} \end{gathered}$

Then we substract 101.6° from both sides:

$\begin{gathered} B+101.60^{\circ}-101.60^{\circ}=180^{\circ}-101.60^{\circ} \\ B=78.40^{\circ} \end{gathered}$

Then the law of sines looks like this:

$\frac{8}{\sin66.60^{\circ}}=\frac{b}{\sin78.40^{\circ}}=\frac{5}{\sin35^{\circ}}$

So we have the following equation for b:

$\frac{b}{\sin78.40^{\circ}}=\frac{5}{\sin35^{\circ}}$

We multiply both sides by sin(78.40°) and we find b:

$\begin{gathered} \frac{b}{\sin78.40^{\circ}}\cdot\sin 78.40^{\circ}=\frac{5}{\sin35^{\circ}}\sin 78.40^{\circ} \\ b=\frac{5}{\sin35^{\circ}}\sin 78.40^{\circ}=8.54 \end{gathered}$

Now let's do the same for the obtuse value of A: 113.40°. The sum of the internal angles is:

$\begin{gathered} A+B+C=180^{\circ} \\ 113.40^{\circ}+B+35^{\circ}=180^{\circ} \\ B+148.40^{\circ}=180^{\circ} \end{gathered}$

We substract 148.40° from both sides:

$\begin{gathered} B+148.40^{\circ}-148.40^{\circ}=180^{\circ}-148.40^{\circ} \\ B=31.6^{\circ} \end{gathered}$

Then the law of sines is:

$\frac{8}{\sin113.40^{\circ}}=\frac{b}{\sin31.60^{\circ}}=\frac{5}{\sin35^{\circ}}$

Then we get this equation for b:

$\frac{b}{\sin31.60^{\circ}}=\frac{5}{\sin35^{\circ}}$

And we multiply both sides by 31.60°:

$\begin{gathered} \frac{b}{\sin31.60^{\circ}}\cdot\sin 31.60^{\circ}=\frac{5}{\sin35^{\circ}}\cdot\sin 31.60^{\circ} \\ b=4.57 \end{gathered}$Answers

The answers for triangle 1 are:

A = 66.60°

B = 78.40°

b = 8.54

The answers for triangle 2 are:

A = 113.40°

B = 31.60°

b = 4.57