How do you solve the triangle given C=80^circ, a=9, b=2?

Dec 18, 2016

Triangle is $a = 9$, $b = 2$, $c = 8.874$, $A = {87.17}^{o}$, $B = {12.83}^{o}$ and $C = {80}^{o}$.

Explanation:

Solving a triangle means identifying length of all the three sides as well as measures of all three angles. This is generally done using Law of sines, which is $\frac{a}{\sin} A = \frac{b}{\sin} B = \frac{c}{\sin} C$ and Law of cosines, according to which ${b}^{2} = {a}^{2} + {c}^{2} - 2 a c \cos B$, ${c}^{2} = {a}^{2} + {b}^{2} - 2 a b \cos C$ and ${a}^{2} = {b}^{2} + {c}^{2} - 2 b c \cos A$

Here, we are given $a = 9$, $b = 2$ and $C = {80}^{o}$.

We can use ${c}^{2} = {9}^{2} + {2}^{2} - 2 \times 9 \times 2 \times \cos {80}^{o}$

= $81 + 4 - 36 \times 0.17365 = 85 - 6.2514 = 78.7486$

Hence $c = \sqrt{78.7486} = 8.874$

Now using ^^Law of sines**

$\frac{9}{\sin} A = \frac{2}{\sin} B = \frac{8.874}{\sin {80}^{o}} = \frac{8.874}{0.9848} = 9.011$

Hence $\sin A = \frac{9}{9.011} = 0.9988$ and $A = {87.17}^{o}$

and $\sin B = \frac{2}{9.011} = 0.222$ and $B = {12.83}^{o}$

Hence, triangle is $a = 9$, $b = 2$, $c = 8.874$, $A = {87.17}^{o}$, $B = {12.83}^{o}$ and $C = {80}^{o}$.