Properties and Solution of Triangles
2.0 Cosine Rule
2.0 Cosine Rule
This law relates the sides of a triangle to the cosine of one of its angles. Mathemetically, it is stated as $$\begin{equation} \begin{aligned} \cos A = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}} \\ {a^2} = {b^2} + {c^2} - 2bc\cos A \\\end{aligned} \end{equation} $$
Similarly $$\begin{equation} \begin{aligned} \cos B = \frac{{{a^2} + {c^2} - {b^2}}}{{2ac}} \\ \cos C = \frac{{{a^2} + {b^2} - {c^2}}}{{2ab}} \\\end{aligned} \end{equation} $$
Proof: Consider a triangle $ABC$, draw an altitude $h$ from the vertex $A$ which intersects the side $BC$ at a point $D$ as shown in the figure.
In the triangle $ABD$, using pythagoras theorem, we can write as $$\begin{equation} \begin{aligned} {c^2} = {h^2} + {(a - x)^2} \\ {c^2} = {h^2} + {a^2} + {x^2} - 2ax \\\end{aligned} \end{equation} $$
Similarly in triangle $ADC$, using pythagoras theorem, we can write as
$${b^2} = {h^2} + {x^2}$$
By using the definition of cosine angle in triangle $ADC$, we can write as $$\begin{equation} \begin{aligned} \cos C = \frac{x}{b} \\ x = b\cos C \\\end{aligned} \end{equation} $$
Now eliminate $x$ and $h$, $$\begin{equation} \begin{aligned} {c^2} = {b^2} - {x^2} + {a^2} + {x^2} - 2ax \\ {c^2} = {b^2} + {a^2} - 2ax \\ {c^2} = {b^2} + {a^2} - 2ab\cos C \\ \cos C = \frac{{{b^2} + {a^2} - {c^2}}}{{2ab}} \\\end{aligned} \end{equation} $$
Question 2. In a triangle $ABC$, if $a=13$, $b=8$ and $c=7$, then find the value of $\sin A$.
Solution: $$\begin{equation} \begin{aligned} \cos A = \frac{{{b^2} + {c^2} - {a^2}}}{{2cb}} \\ \cos A = \frac{{{8^2} + {7^2} - {{13}^2}}}{{2 \times 7 \times 8}} \\ \cos A = \frac{{64 + 49 - 169}}{{2 \times 7 \times 8}} \\ \cos A = \frac{{ - 56}}{{2 \times 56}} \\ \cos A = - \frac{1}{2} \\ A = \frac{{2\pi }}{3} \\ \sin A = \sin \frac{{2\pi }}{3} \\ \sin A = \frac{{\sqrt 3 }}{2} \\\end{aligned} \end{equation} $$
Question 3. If in a triangle $ABC$, $\angle A = {60^{^0}}$, then find the value of $$(1 + \frac{a}{c} + \frac{b}{c})(1 + \frac{c}{b} - \frac{a}{b})$$
Solution: $$\begin{equation} \begin{aligned} (1 + \frac{a}{c} + \frac{b}{c})(1 + \frac{c}{b} - \frac{a}{b}) = (\frac{{c + a + b}}{c})(\frac{{b + c - a}}{b}) \\ (\frac{{{{(b + c)}^2} - {a^2}}}{{bc}}) \\ \frac{{{b^2} + {c^2} - {a^2} + 2bc}}{{bc}} \\ \frac{{{b^2} + {c^2} - {a^2}}}{{bc}} + 2 \\\end{aligned} \end{equation} $$
$$2(\frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}) + 2$$
Now using cosine rule $$\cos A = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}$$ we can write it as
$$\begin{equation} \begin{aligned} 2\cos A + 2 \\ 2\cos {60^{^{^0}}} + 2 \\ 2 \times \frac{1}{2} + 2 \\ 1 + 2 \\ 3 \\\end{aligned} \end{equation} $$
Question 4. If $a$, $b$ and $A$ are given in a triangle and ${c_1}$ and ${c_2}$ are the possible values of the third side, prove that $${c_1}^2 + {c_2}^2 - 2{c_1}{c_2}\cos 2A = 4{a^2}{\cos ^2}A$$
Solution: We know that $$\cos A = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}$$
$${c^2} - 2bc\cos A + {b^2} - {a^2} = 0$$
Above equation is quadratic in $c$ so by using the sum and products of roots, we have$$\begin{equation} \begin{aligned} {c_1} + {c_2} = 2b\cos A \\ {c_1}{c_2} = {b^2} - {a^2} \\ {c_1}^2 + {c_2}^2 - 2{c_1}{c_2}\cos 2A = {({c_1} + {c_2})^2} - 2{c_1}{c_2}(1 + \cos 2A) \\\end{aligned} \end{equation} $$
$$\begin{equation} \begin{aligned} = 4{b^2}{\cos ^2}A - 2({b^2} - {a^2})2{\cos ^2}A \\ = 4{a^2}{\cos ^2}A \\\end{aligned} \end{equation} $$
Question 5. For the triangle $ABC$, it is given that $\cos A + \cos B + \cos C = \frac{3}{2}$, then the triangle is equilateral.
Solution: Let $ABC$ be a triangle in which $a$, $b$ and $c$ are the sides of a triangle.
given that $\cos A + \cos B + \cos C = \frac{3}{2}$
so by using the cosine rule, we can write as $$\begin{equation} \begin{aligned} (\frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}) + (\frac{{{c^2} + {a^2} - {b^2}}}{{2ac}}) + (\frac{{{a^2} + {b^2} - {c^2}}}{{2ab}}) = \frac{3}{2} \\ \frac{{{b^2}a + {c^2}a - {a^3} + {c^2}b + {a^2}b - {b^3} + {a^2}c + {b^2}c - {c^3}}}{{2abc}} = \frac{3}{2} \\ {b^2}a + {c^2}a - {a^3} + {c^2}b + {a^2}b - {b^3} + {a^2}c + {b^2}c - {c^3} = 3abc \\ \\\end{aligned} \end{equation} $$
Add $-6abc$ on both sides we get,
$$\begin{equation} \begin{aligned} {b^2}a + {c^2}a - {a^3} + {c^2}b + {a^2}b - {b^3} + {a^2}c + {b^2}c - {c^3} - 6abc = 3abc - 6abc \\ {b^2}a + {c^2}a - {a^3} + {c^2}b + {a^2}b - {b^3} + {a^2}c + {b^2}c - {c^3} - 2abc - 2abc - 2abc = 3abc - 6abc \\ {b^2}a + {c^2}a - 2abc + {c^2}b + {a^2}b - 2abc + {a^2}c + {b^2}c - 2abc = - 3abc + {a^3} + {b^3} + {c^3} \\ ({b^2}a + {c^2}a - 2abc) + ({c^2}b + {a^2}b - 2abc) + ({a^2}c + {b^2}c - 2abc) = - 3abc + {a^3} + {b^3} + {c^3} \\ a({b^2} + {c^2} - 2bc) + b({c^2} + {a^2} - 2ac) + c({a^2} + {b^2} - 2ab) = {a^3} + {b^3} + {c^3} - 3abc \\ a{(b - c)^2} + b{(c - a)^2} + c{(a - b)^2} = {a^3} + {b^3} + {c^3} - 3abc \\ a{(b - c)^2} + b{(c - a)^2} + c{(a - b)^2} = (a + b + c)\frac{1}{2}[{(a - b)^2} + {(b - c)^2} + {(c - a)^2}] \\ \\\end{aligned} \end{equation} $$
This gives $$(a + b - c){(a - b)^2} + (b + c - a){(b - c)^2} + (c + a - b){(c - a)^2} = 0$$
but we know that $$(a + b - c) > 0;(b + c - a) > 0;(c + a - b) > 0$$
on LHS side, we have coefficient multiplied by square of a number. Each coefficient is positive and hence for the sum to be zero, each term separately must be zero, therefore, $$\begin{equation} \begin{aligned} (a + b - c){(a - b)^2} = 0 \\ (b + c - a){(b - c)^2} = 0 \\ (c + a - b){(c - a)^2} = 0 \\\end{aligned} \end{equation} $$
which implies that $$a=b=c$$
Therefore, $ABC$ is an equilateral triangle.
Question 6. If in a triangle$ABC$, $a=6$, $b=3$ and $\cos (A - B) = \frac{4}{5}$. Find the area of triangle.
Solution: Given $$\cos (A - B) = \frac{4}{5}$$
Using $$\cos 2A = \frac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}}$$
we can write as $$\cos (A - B) = \frac{{1 - {{\tan }^2}(\frac{{A - B}}{2})}}{{1 + {{\tan }^2}(\frac{{A - B}}{2})}}$$
$$\frac{{1 - {{\tan }^2}(\frac{{A - B}}{2})}}{{1 + {{\tan }^2}(\frac{{A - B}}{2})}} = \frac{4}{5}$$
By componendo and dividendo, we get $$\begin{equation} \begin{aligned} {\tan ^2}(\frac{{A - B}}{2}) = \frac{1}{9} \\ \tan (\frac{{A - B}}{2}) = \frac{1}{3} \\\end{aligned} \end{equation} ...since A>B $$
$$\begin{equation} \begin{aligned} \tan (\frac{{A - B}}{2}) = \frac{{a - b}}{{a + b}}\cot \frac{C}{2} \\ \frac{1}{3} = \frac{{6 - 3}}{{6 + 3}}\cot \frac{C}{2} \\ \cot \frac{C}{2} = 1 \\ C = \frac{\pi }{2} \\\end{aligned} \end{equation} $$
Area of triangle is given as $$\begin{equation} \begin{aligned} \frac{1}{2}ab\sin C \\ \frac{1}{2} \times 6 \times 3 \times \sin \frac{\pi }{2} \\ = 9 \\\end{aligned} \end{equation} $$
So, area of triangle is $9$ square units.
