# How do you determine the number of possible triangles and find the measure of the three angles given a=8, b=10, mangleA=20?

Oct 31, 2017

$A = {20}^{\circ} , {B}_{1} = {25}^{\circ} 19 ' , {C}_{1} = {134}^{\circ} 41 '$ and

$A = {20}^{\circ} , {B}_{2} = {154}^{\circ} 41 ' ' , {C}_{2} = {5}^{\circ} 19 '$

#### 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,

$h = 10 \sin {20}^{\circ} \approx 3.42$, since $3.42 < 8 < 10$ we have

$h < a < b$ so we are looking for two solutions. Hence,

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

${8}^{2} = {10}^{2} + {c}^{2} - 2 \left(10\right) \left(c\right) \cos {20}^{\circ}$

$64 = 100 + {c}^{2} - \left(20 \cos {20}^{\circ}\right) c$

$0 = {c}^{2} - \left(20 \cos {20}^{\circ}\right) c + 36$

$c = \frac{\left(20 \cos {20}^{\circ}\right) \pm \sqrt{{\left(- 20 \cos {20}^{\circ}\right)}^{2} - 4 \left(1\right) \left(36\right)}}{2}$

$c = \frac{\left(20 \cos {20}^{\circ}\right) + \sqrt{{\left(- 20 \cos {20}^{\circ}\right)}^{2} - 144}}{2}$ or

$c = \frac{\left(20 \cos {20}^{\circ}\right) - \sqrt{{\left(- 20 \cos {20}^{\circ}\right)}^{2} - 144}}{2}$

$\therefore {c}_{1} \approx 16.63 \mathmr{and} {c}_{2} \approx 2.16$

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

${B}_{1} = {\cos}^{-} 1 \left[\frac{{8}^{2} + {c}_{1}^{2} - {10}^{2}}{2 \cdot {c}_{1} \cdot 8}\right] = {25}^{\circ} 19 '$

and therefore

${C}_{1} = {180}^{\circ} - {20}^{\circ} - {25}^{\circ} 19 ' = {134}^{\circ} 41 '$

${B}_{2} = {\cos}^{-} 1 \left[\frac{{8}^{2} + {c}_{2}^{2} - {10}^{2}}{2 \cdot {c}_{2} \cdot 8}\right] = {154}^{\circ} 41 '$

and therefore

${C}_{2} = {180}^{\circ} - {20}^{\circ} - {154}^{\circ} 41 ' = {5}^{\circ} 19 '$