Search any question & find its solution
Question:
Answered & Verified by Expert
If the angles of a $\triangle A B C$ are in $\mathrm{AP}$, then
Options:
Solution:
2564 Upvotes
Verified Answer
The correct answer is:
$b^2=a^2+c^2-a c$
Let the angle of $\triangle A B C$, be $(a-d), a,(a+d)$
So, $\quad a-d+a+a+d=180^{\circ} \Rightarrow a=60^{\circ}$
$\begin{aligned} \cos B & =\frac{a^2+c^2-b^2}{2 a c} \\ \Rightarrow \quad \frac{1}{2} & =\frac{a^2+c^2-b^2}{2 a c} \Rightarrow b^2=a^2+c^2-a c\end{aligned}$
So, $\quad a-d+a+a+d=180^{\circ} \Rightarrow a=60^{\circ}$
$\begin{aligned} \cos B & =\frac{a^2+c^2-b^2}{2 a c} \\ \Rightarrow \quad \frac{1}{2} & =\frac{a^2+c^2-b^2}{2 a c} \Rightarrow b^2=a^2+c^2-a c\end{aligned}$
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.