Trigonometric Ratios II - SS2 Mathematics Past Questions and Answers - page 1

Using the sine rule, \(\frac{a}{\sin A} = \frac{b}{\sin B}\)

\[\frac{3.80}{\sin 62} = \frac{b}{\sin 26}\]

\(b = \ \frac{\sin 26 \times 3.80}{\sin 62} = \ \frac{0.4384 \times 3.8}{0.8829} = 1.89cm\) 

Using the cosine rule, \(c^{2} = a^{2} + b^{2} - 2ab\cos C\)

\[c^{2} = {5.3}^{2} + {8.9}^{2} - 2(5.3)(8.9)\cos 55\]

\[c^{2} = 28.09 + 79.21 - 2(5.3)(8.9)(0.5736)\]

\[c^{2} = 28.09 + 79.21 - 54.11\]

\(c = \sqrt{53.19} = 7.3cm\)

First, calculate the \(\angle C\)

\[\angle C = 180{^\circ} - (\angle A + \angle B)\]

\[\angle C = 180{^\circ} - (145{^\circ} + 23{^\circ})\]

\[\angle C = 180{^\circ} - 168{^\circ}\]

\[\angle C = 12{^\circ}\]

From the Sine Rule, \(\frac{a}{\sin A} = \frac{b}{\sin B}\)

\[\frac{a}{\sin 145} = \frac{5.02}{\sin 23}\]

\[\frac{a}{0.5736} = \frac{5.02}{0.3907}\]

\[a = \frac{5.02 \times 0.5736}{0.3907} \approx 7.37cm\]

From the Sine Rule, \(\frac{b}{\sin B} = \frac{c}{\sin C}\)

\[\frac{5.02}{\sin 23} = \frac{c}{\sin 12}\]

\[\frac{5.02}{0.3907} = \frac{c}{0.2079}\]

\[c = \frac{5.02 \times 0.2079}{0.3907} \approx 2.67cm\]

Using the cosine rule, \(c^{2} = a^{2} + b^{2} - 2ab\cos C\)

\[c^{2} = {2.25}^{2} + {6.24}^{2} - 2(2.25)(6.24)\cos 78\]

\[c^{2} = 5.0625 + 38.9376 - 2(2.25)(6.24)(0.2079)\]

\[c^{2} = 5.0625 + 38.9376 - 5.837832\]

\[c^{2} = 38.162268\]

\[c = \sqrt{38.162268} = 6.18cm\]

\(distance\ bewteen\ boats\ A\ and\ B = distance\ of\ B\ from\ cliff - disatnce\ of\ A\ from\ cliff\)

\[disatnce\ of\ A\ from\ cliff:\]

\[\tan{42{^\circ}} = \frac{21}{OA}\]

\[OA = \frac{21}{\tan{42{^\circ}}} = \frac{21}{0.9004} = 23.32m\]

\[disatnce\ of\ B\ from\ cliff:\]

\[\tan{25{^\circ}} = \frac{21}{OB}\]

\[OB = \frac{21}{\tan{25{^\circ}}} = \frac{21}{0.4663} = 45.04m\]

\[distance\ bewteen\ boats\ A\ and\ B = distance\ of\ B\ from\ cliff - disatnce\ of\ A\ from\ cliff\]

\[distance\ bewteen\ boats\ A\ and\ B = 45.04 - 23.32 = \ 21.72m\]

\(\angle O = 23{^\circ} + (180{^\circ} - 113{^\circ})\)

\[\angle O = 23{^\circ} + 67{^\circ}\]

\[\angle O = 90{^\circ}\]

Using the cosine rule,

\[o^{2} = p^{2} + q^{2} - 2pq\cos O\]

\[o^{2} = 4^{2} + 3^{2} - 2(4)(3)\cos 90\]

\[o^{2} = 16 + 9 - 0\]

\[o^{2} = 25\]

\[o = \sqrt{25} = \ 5km\]