In a triangle PQR, R=2.

Question

In a triangle $PQR$, $\angle R=\dfrac\pi2$. If $	an\dfrac P2$ and $	an\dfrac Q2$ are the roots of $ax^2+bx+c=0$, $a
e0$, then

Accepted Answer

$\tan\dfrac P2+\tan\dfrac Q2=-\dfrac ba$, $\tan\dfrac P2\tan\dfrac Q2=\dfrac ca$. Since $P+Q=\dfrac\pi2$: $\tan\left(\dfrac P2+\dfrac Q2\right)=1$.

$$\frac{-b/a}{1-c/a}=1 \implies -b=a-c \implies c=a+b.$$

Solutions · 1