Show that $\sin 105{}^\circ = \frac{\sqrt{6} + \sqrt{2}}{4}$.[3] The diagram below shows tria

Question

Show that $\sin 105{}^\circ = \frac{\sqrt{6} + \sqrt{2}}{4}$.[3] The diagram below shows triangle $ABC$, such that $AB = 4$ cm, $\angle ABC = 105{}^\circ$ and the area of triangle $ABC$ is $10 + 2\sqrt{3}$ cm${}^2$.Find $BC$ in the form $\sqrt{2}(a\sqrt{3} + b)$, where $a$ and $b$ are integers.[3]

Accepted Answer

(a) $\sin105^\circ=\dfrac{\sqrt{6}+\sqrt{2}}{4}$ (b) $BC=\sqrt{2}(4\sqrt{3}-2)\text{ cm}$

2025 DSS PRELIMS P1 Q6

2025 DSS PRELIMS P1 Q6

Solution:

Solution locked