IIT-JEE 1980Question 1
$ABC$ is a triangle. $D$ is the middle point of $BC$. If $AD$ is perpendicular to $AC$, then prove that $$\cos A\cos C=\frac{2\left(c^{2}-a^{2}\right)}{3ac}$$
Trigonometric Ratio & Identities
Answerproved
Solution
Concept. Put $A$ at the origin, $AC$ on the $x$-axis and $AD$ on the $y$-axis; then $B$ is fixed by $D$ being the midpoint of $BC$. Formulas used. (1) $\cos(\text{angle between }\vec u,\vec v)=\dfrac{\vec u\cdot\vec v}{|\vec u||\vec v|}$Step 1 - Coordinates. $A(0,0)$, $C(b,0)$, $D(0,d)$, so $B=2D-C=(-b,2d)$. Then $$c^2=AB^2=b^2+4d^2,\qquad a^2=BC^2=4b^2+4d^2,\qquad c^2-a^2=-3b^2.$$ Step 2 - $\cos A$. By (1) with $\vec{AB}=(-b,2d)$, $\vec{AC}=(b,0)$: $\cos A=\dfrac{-b^2}{cb}=-\dfrac bc$. Step 3 - $\cos C$. With $\vec{CA}=(-b,0)$, $\vec{CB}=(-2b,2d)$: $\cos C=\dfrac{2b^2}{ba}=\dfrac{2b}{a}$. Step 4 - Product. $$\cos A\cos C=-\frac{2b^2}{ac}=\frac{2(c^2-a^2)}{3ac}.$$ Hence proved.