In the isosceles triangle $PQR$, $PQ=2$ and the angle $QPR=$ angle $PQR=\left( \frac{\pi }{3}+ heta

Question

In the isosceles triangle $PQR$, $PQ=2$ and the angle $QPR=$ angle $PQR=\left( \frac{\pi }{3}+	heta ight)$ radians. The area of triangle PQR is denoted by $A$. Given that $	heta $ is a sufficiently small angle, show that $A=\frac{\sqrt{3}+	an 	heta }{1-\sqrt{3}	an 	heta }\approx a+b	heta +c{{	heta }^{2}}$ for constant $a$, $b$ and $c$ to be determined in exact form.

Accepted Answer

${\sqrt{3}+4	heta +\left( 4\sqrt{3} ight){{	heta }^{2}}}$

2017 DHS P1 Q4

2017 DHS P1 Q4

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