In the triangle $ABC$, $AB=2$, angle $ABC= heta $ radians and angle $BAC=\frac{\pi }{4}$ radians. S

Question

In the triangle $ABC$, $AB=2$, angle $ABC=	heta $ radians and angle $BAC=\frac{\pi }{4}$ radians. Show that $BC=\frac{2}{\sin 	heta +\cos 	heta }$. [3] Given that $	heta $ is a sufficiently small angle, show that $BC\approx p+q	heta +r{{	heta }^{2}}$, for constants $p$,$q$ and $r$ to de determined. [3]

Accepted Answer

$p=2,\; q=-2,\; r=3$

2022 ACJC J2 MYE Q1

2022 ACJC J2 MYE Q1

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