Question

In △ABC, m∠A=15 °, a=10 , and b=11 . Find c to the nearest tenth.

Answer 1

Answer:

The answer is:

$\bold{c\approx 20.2\ units}$

Step-by-step explanation:

Given:

In △ABC:

m∠A=15°

a=10 and

b=11

To find:

c = ?

Solution:

We can use cosine rule here to find the value of third side c.

Formula for cosine rule:

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

Where  

a is the side opposite to $\angle A$

b is the side opposite to $\angle B$

c is the side opposite to $\angle C$

Putting all the values.

$cos 15^\circ = \dfrac{11^{2}+c^{2}-10^{2}}{2\times 11 \times c}\\\Rightarrow 0.96 = \dfrac{121+c^{2}-100}{22c}\\\Rightarrow 0.96 \times 22c= 121+c^{2}-100\\\Rightarrow 21.25 c= 21+c^{2}\\\Rightarrow c^{2}-21.25c+21=0\\\\\text{solving the quadratic equation:}\\\\c = \dfrac{21.25+\sqrt{21.25^2-4 \times 1 \times 21}}{2}\\c = \dfrac{21.25+\sqrt{367.56}}{2}\\c = \dfrac{21.25+19.17}{2}\\c \approx 20.2\ units$

The answer is:

$\bold{c\approx 20.2\ units}$