isi-entrance 2005 Q1

isi-entrance · India · solved Sine and Cosine Rules Determine an angle or side from a trigonometric identity/equation
In a right angle triangle with sides $a < b < c$, where $\angle ACB = \theta$ is the smallest angle, show that $\sin^2\theta - \sqrt{5}\sin\theta + 1 = 0$, given that $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2}$ (i.e., the reciprocals of the sides also form a right triangle).
From Pythagoras: $a^2 + b^2 = c^2$ ... (I)
Also given: $(1/b)^2 + (1/c)^2 = (1/a)^2 \Rightarrow (b^2 + c^2)a^2 = b^2c^2$ ... (II)
Since $\sin\theta = a/c$, from (I): $b^2 = c^2\cos^2\theta$.
Substituting into (II): $(\cos^2\theta + 1)\sin^2\theta = \cos^2\theta$ $\Rightarrow (2 - \sin^2\theta)\sin^2\theta = \sin^2\theta(1 - \sin^2\theta)$ $\Rightarrow \sin^4\theta - 3\sin^2\theta + 1 = 0$ $\Rightarrow (\sin^2\theta + 1)^2 = 5\sin^2\theta$ $\Rightarrow \sin^2\theta - \sqrt{5}\sin\theta + 1 = 0$ 
In a right angle triangle with sides $a < b < c$, where $\angle ACB = \theta$ is the smallest angle, show that $\sin^2\theta - \sqrt{5}\sin\theta + 1 = 0$, given that $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2}$ (i.e., the reciprocals of the sides also form a right triangle).
Paper Questions
Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10