In the diagram above, $OAB$ is an isosceles triangle where $OA=OB=r$ cm. It is given that t

Question

In the diagram above, $OAB$ is an isosceles triangle where $OA=OB=r$ cm. It is given that the length $AB=2$ cm and angle $AOB=\frac{\pi }{3}-\theta $ radians. Using cosine rule, show that ${{r}^{2}}=\frac{4}{2-\sqrt{3}\sin \theta -\cos \theta }$.[3] Given that $\theta $ is a sufficiently small angle, show that $r\approx 2+a\theta +b{{\theta }^{2}}$ where $a$ and $b$ are constants to be determined.[4]

Accepted Answer

(i) $r^2=\frac{4}{2-\sqrt3\sin\theta-\cos\theta}$ $(\text{shown})$ (ii) $r\approx2+\sqrt3\theta+\frac74\theta^2$, so $a=\sqrt3$ and $b=\frac74$

2024 TMJC P1 Q4

2024 TMJC P1 Q4

Video Solution:

Solution:

Solution locked