# How do you determine the number of possible triangles and find the measure of the three angles given b=12, c=10, mangleB=49?

Oct 31, 2017

$\therefore A = {92}^{\circ} 02 ' , B = {49}^{\circ} , \mathmr{and} C = {38}^{\circ} 58 '$

#### Explanation:

Since the given information is for a SSA triangle it is the ambiguous case. In the ambiguous case we first find the height by using the formula $h = b \sin A$.

Note that A is the given angle and its side is always a so the other side will be b.

So if $A < {90}^{\circ}$ and if

1. $h < a < b$ then then there are two solutions or two triangles.

2. $h < b < a$ then there is one solution or one triangle.

3. $a < h < b$ then there is no solution or no triangle.

If $A \ge {90}^{\circ}$ and if

1. $a > b$ then there is one solution or one triangle.

2. $a \le b$ there is no solution

Now let's use the Law of Cosine ${a}^{2} = {b}^{2} + {c}^{2} - 2 b c \cos A$ and the

quadratic formula $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$to figure out our solutions.

That is,

$h = 10 \sin {49}^{\circ} \approx 7.55$ and since $7.55 < 10 < 12$ we have

$h < b < a$ so we are looking for one solution. Hence,

${b}^{2} = {a}^{2} + {c}^{2} - 2 a c \cos B$

${12}^{2} = {a}^{2} + {10}^{2} - 2 \left(10\right) \left(a\right) \cos {49}^{\circ}$

$144 = {a}^{2} + 100 - \left(20 \cos {49}^{\circ}\right) a$

$0 = {a}^{2} - \left(20 \cos {49}^{\circ}\right) a - 44$

$a = \frac{\left(20 \cos {49}^{\circ}\right) \pm \sqrt{{\left(20 \cos {49}^{\circ}\right)}^{2} - 4 \left(1\right) \left(- 44\right)}}{2}$

$a = \frac{\left(20 \cos {49}^{\circ}\right) + \sqrt{{\left(20 \cos {49}^{\circ}\right)}^{2} + 176}}{2}$ or

$a = \frac{\left(20 \cos {49}^{\circ}\right) - \sqrt{{\left(20 \cos {49}^{\circ}\right)}^{2} + 176}}{2}$

a~~15.89 or cancel(a~~-2.77

$\therefore a \approx 15.89$

To find the measure of angle A we use the law of cosine and solve for A. That is,

$A = {\cos}^{-} 1 \left[\frac{{12}^{2} + {10}^{2} - {a}^{2}}{2 \cdot 12 \cdot 10}\right] = {92}^{\circ} 02 '$

and therefore

$C = {180}^{\circ} - {49}^{\circ} - {92}^{\circ} 02 ' = {38}^{\circ} 58 '$