In the triangle $PQR$, angle $PQR=\frac{\pi }{3}$ radians and angle $PRQ=\frac{\pi }{3}- heta $ rad

Question

In the triangle $PQR$, angle $PQR=\frac{\pi }{3}$ radians and angle $PRQ=\frac{\pi }{3}-	heta $ radians, where $	heta $ is small enough for ${{	heta }^{3}}$ and higher powers of $	heta $ to be negligible. Show that $\frac{PR}{PQ}\approx 1+\alpha 	heta +\beta {{	heta }^{2}}$, where$\alpha $ and $\beta $ are exact real constants to be determined.[5]

Accepted Answer

$\alpha=\frac1{\sqrt3}$, $\beta=\frac56$

2023 HCI P1 Q2

2023 HCI P1 Q2

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