Question

In a triangle ABC, $a = 8cm,b = 10cm$ and $c = 12cm$. The relation between the angles of the triangle is
A. $C = A + B$
B. $C = 2B$
C. $C = 2A$
D. $C = 3A$

Answer 1

Hint: In the above question we have been given the sides of the triangle and their measurements. We will first draw the diagram of the triangle with the given data and then solve it.
We will us the formula to find the value of
$\cos C$ .
The formula for the value of $\cos C$ is
 $\dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}$ .
Similarly we will also calculate the value of $\cos A$ with its formula.

Complete step-by-step solution:
It is given that in $\Delta ABC$, we have
$a = 8cm,b = 10cm$ and
$c = 12cm$.
Let us draw the diagram of the above:

Now we will calculate the value of $\cos C$ i.e.
$\dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}$ .
By putting the values we have :
$\dfrac{{{8^2} + {{10}^2} - {{12}^2}}}{{2 \times 8 \times 10}}$
On simplifying the values we have :
$\dfrac{{64 + 100 - 144}}{{160}}$
It gives us value
$\dfrac{{20}}{{160}} = \dfrac{1}{8}$
Similarly we will find the value of
$\cos C$.
The formula is
$\cos C = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}$ .
We will put the value and we have:
$\dfrac{{{{10}^2} + {{12}^2} - {8^2}}}{{2 \times 10 \times 12}}$
On further simplifying it, we can write it as
$\dfrac{{100 + 144 - 64}}{{240}}$
It gives us the value
$\dfrac{{180}}{{240}} = \dfrac{3}{4}$ .
Now we ill use the trigonometric identity i.e.
$\cos 2A = 2{\cos ^2}A - 1$
By putting the value form the above we can write it as
$\cos 2A = 2{\left( {\dfrac{3}{4}} \right)^2} - 1$
On further simplifying it gives us the value
 $2 \times \dfrac{9}{{16}} - 1 = \dfrac{9}{8} - 1$
It gives us value
$\dfrac{{9 - 8}}{8} = \dfrac{1}{8}$
Now we can say that the value of $\cos 2A$ is equal to the value of $\cos C$.
It can be written as
$\cos C = \cos 2A$
We can cancel out the cosine term , so it gives us
$C = 2A$
Hence the correct option is (c) $C = 2A$

Note: We should also remember the formula of cosine difference formula i.e.
$\cos (A - B)$ can be written as
$\cos A\cos B + \operatorname{Sin} A\sin B$ .
If we have to find the value of $\sin C$ in the above solution, then the formula is
$\sin C = \sqrt {1 - {{\cos }^2}C} $
So it gives us
$\sqrt {1 - {{\left( {\dfrac{1}{8}} \right)}^2}} $.
On simplifying it gives the value
 $\sqrt {1 - \dfrac{1}{{64}}} = \sqrt {\dfrac{{64 - 1}}{{64}}} $
The value is
$\sqrt {\dfrac{{63}}{{64}}} $ .
It can also be written as
$\dfrac{{3\sqrt 7 }}{8}$ .