Question

In a $∆ABC,a=8cm,b=10cm,c=12cm$.

Then, the relation between angles of the triangle is

• A. $C=A+B$
• B. $C=2B$
• C. $C=2A$
• D. $C=3A$

Answer 1

The correct option is C

$C=2A$

Explanation for the correct option:

Step 1: Find the $\sin C$:

Given $a=8cm,b=10cm,c=12cm$

We know, $\cos C=\frac{(a^2+b^2–c^2)}{2ab}$

Substituting the given values of $a,bandc$ we get,

$\begin{array}{rcl}\cos C& =& \frac{(8^2+{10}^2–{12}^2)}{2\times 8\times 10}\\ & =& \frac{(64+100-144)}{160}\\ & =& \frac{20}{160}\\ & =& \frac{1}{8}\end{array}$

So,

$$\begin{gathered} \sin C=\sqrt{(1–\frac{1}{64})}\left[\because {\sin }^{2}A+{\cos }^{2}A=1\right] \\ =\sqrt{(63/64)} \\ =\frac{3\sqrt{7}}{8} \end{gathered}$$

Step 2: Find the $\sin A$:

Now similarly, $\cos A=\frac{(b^2+c^2–a^2)}{2bc}$

Putting in the values we get,

$$\begin{gathered} \cos A=\frac{({10}^2+{12}^2-8^2)}{2\times 10\times 12} \\ =\frac{9}{12} \\ =\frac{3}{4} \end{gathered}$$

Therefore,

$$\begin{gathered} \sin A=\sqrt{(1–9/16)} \\ =\frac{\sqrt{7}}{4} \end{gathered}$$

Now,

$$\begin{gathered} \frac{\cos C}{\cos A}=\frac{1}{8}\times \frac{4}{3} \\ =\frac{1}{6} \end{gathered}$$

Step 3: Finding the relation between the angles:

We know,

$\cos (C-A)=\cos C\cos A+\sin C\sin A$

Substituting the values we get,

$\begin{array}{rcl}\cos \left(C-A\right)& =& \left(\frac{1}{8}\right)\times \left(\frac{3}{4}\right)+\frac{3\sqrt{7}}{8}\times \left(\frac{\sqrt{7}}{4}\right)\\ & =& \frac{3}{32}+\frac{21}{32}\\ & =& \frac{24}{32}\\ & =& \frac{3}{4}\\ & =& \cos A\end{array}$

Now,

$$\begin{gathered} \cos (C-A)=\cos A \\ (C-A)=A \\ C=2A \end{gathered}$$

Hence, the correct option is (C).