(a) In the triangle $ABC$ as shown below, $BC=3$, angle $BAC=\frac{\pi }{3}+ heta $ radians and ang

Question

(a) In the triangle $ABC$ as shown below, $BC=3$, angle $BAC=\frac{\pi }{3}+	heta $ radians and angle $ACB=\frac{\pi }{2}$ radians. Show that $AC=\frac{3\left( 1-\sqrt{3}	an 	heta ight)}{\sqrt{3}+	an 	heta }$. [3] Given that $	heta $ is a sufficiently small angle, deduce that $AC\approx \sqrt{3}+a	heta $, where $a$ is a constant to be determined. [3]

Accepted Answer

$a=-4$

2017 PJC Promo Q9

2017 PJC Promo Q9

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