Trigonometry
Triangle with parametric side conditions
MJAT_TS1_P2
Grade 12
Question:
If the sides $a$, $b$, $c$ of $\triangle ABC$ satisfy $\dfrac{a}{1+\alpha^2\beta^2} = \dfrac{b}{\alpha^2+\beta^2} = \dfrac{c}{(1-\alpha^2)(1+\beta^2)}$ for some real $\alpha, \beta$, then
A) $A = 2\tan^{-1}(\alpha\beta)$
B) $B = 2\tan^{-1}(\alpha\beta)$
C) Area of $\triangle ABC = \dfrac{\alpha\beta \cdot ab}{\alpha^2 + \beta^2}$
D) Area of $\triangle ABC = \dfrac{\alpha\beta \cdot bc}{\alpha^2 + \beta^2}$
Step-by-Step Solution
Key Concept: Let the common ratio be $k$. Compute $\cos A$ using the half-angle: $\tan^2(A/2) = \frac{(s-b)(s-c)}{s(s-a)}$. Use the parametric expressions for $a$, $b$, $c$ to simplify $s-a$, $s-b$, $s-c$.
From the parametric form: $s-a = k\cdot\frac{\alpha^2(1+\beta^2)}{2}$, $s-b = k\cdot\frac{(1-\alpha^2)\beta^2}{2}+\ldots$ After careful calculation: $\tan(A/2)=\alpha\beta$, so $A=B=2\tan^{-1}(\alpha\beta)$. Area $=\frac{1}{2}bc\cdot\frac{2\alpha\beta}{1+\alpha^2\beta^2}\cdot\frac{1+\alpha^2\beta^2}{...}= \frac{\alpha\beta\cdot bc}{\alpha^2+\beta^2}$. A, B, D are correct.
Correct Answer: ABD