# How do you solve the triangle given a=20, c=24, B=47?

Dec 16, 2016

Triangle is $a = 20$, $b = 17.924$, $c = 24$, $A = {54.7}^{o}$, $B = {47}^{o}$ and $C = {78.3}^{o}$.

#### 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 $a = 20$, $c = 24$ and $B = {47}^{o}$.

We can use ${b}^{2} = {20}^{2} + {24}^{2} - 2 \times 20 \times 24 \times \cos {47}^{o}$

= $400 + 576 - 960 \times 0.682 = 976 - 654.72 = 321.28$

Hence $b = \sqrt{321.28} = 17.924$

Now using ^^Law of sines**

$\frac{20}{\sin} A = \frac{24}{\sin} C = \frac{17.924}{\sin {47}^{o}} = \frac{17.924}{0.7314} = 24.5064$

Hence $\sin A = \frac{20}{24.5064} = 0.8161$ and $A = {54.7}^{o}$

and $\sin C = \frac{24}{24.5064} = 0.9793$ and $C = {78.3}^{o}$

Hence, triangle is $a = 20$, $b = 17.924$, $c = 24$, $A = {54.7}^{o}$, $B = {47}^{o}$ and $C = {78.3}^{o}$.