Search any question & find its solution
Question:
Answered & Verified by Expert
In a $\triangle A B C$, the lengths of the two larger sides are 10 and 9 units, respectively. If the angles are in $\mathrm{AP}$, then the length of the third side can be
Options:
Solution:
1585 Upvotes
Verified Answer
The correct answer is:
$5 \pm \sqrt{6}$
Let A, B and $C$ be the three angles of $\Delta A B C$ and Let $a=10$ and $b=9$ It is given that the angles are in $\mathrm{AP}$. $\therefore 2 \mathrm{~B}=\mathrm{A}+\mathrm{C}$ on adding $\mathrm{B}$ both the sides,
we get
$$
\begin{array}{l}
3 \mathrm{~B}=\mathrm{A}+\mathrm{B}+\mathrm{C} \\
\Rightarrow 3 \mathrm{~B}=180^{\circ} \Rightarrow \mathrm{B}=60^{\circ}
\end{array}
$$
Now, we know $\cos \mathrm{B}=\frac{\mathrm{a}^{2}+\mathrm{c}^{2}-\mathrm{b}^{2}}{2 \mathrm{ac}}$
$$
\begin{array}{l}
\Rightarrow \cos 60^{\circ}=\frac{10^{2}+\mathrm{c}^{2}-9^{2}}{2 \times 10 \times \mathrm{c}} \\
\Rightarrow \frac{1}{2}=\frac{100+\mathrm{c}^{2}-81}{20 \mathrm{c}} \\
\Rightarrow \mathrm{c}^{2}-10 \mathrm{c}+19=0 \Rightarrow \mathrm{c}=5 \pm \sqrt{6}
\end{array}
$$
we get
$$
\begin{array}{l}
3 \mathrm{~B}=\mathrm{A}+\mathrm{B}+\mathrm{C} \\
\Rightarrow 3 \mathrm{~B}=180^{\circ} \Rightarrow \mathrm{B}=60^{\circ}
\end{array}
$$
Now, we know $\cos \mathrm{B}=\frac{\mathrm{a}^{2}+\mathrm{c}^{2}-\mathrm{b}^{2}}{2 \mathrm{ac}}$
$$
\begin{array}{l}
\Rightarrow \cos 60^{\circ}=\frac{10^{2}+\mathrm{c}^{2}-9^{2}}{2 \times 10 \times \mathrm{c}} \\
\Rightarrow \frac{1}{2}=\frac{100+\mathrm{c}^{2}-81}{20 \mathrm{c}} \\
\Rightarrow \mathrm{c}^{2}-10 \mathrm{c}+19=0 \Rightarrow \mathrm{c}=5 \pm \sqrt{6}
\end{array}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.