Trigonometric Ratios II - SS2 Mathematics Past Questions and Answers - page 1
Find the length of the b side of the \(\mathrm{\Delta}ABC\) which have \(\angle A = 62{^\circ},\ \angle B = 26{^\circ}\ \)and\(\ a = 3.80cm\) using the sine rule
1.59cm
2.34cm
1.89cm
2.72cm
Using the sine rule, \(\frac{a}{\sin A} = \frac{b}{\sin B}\)
\[\frac{3.80}{\sin 62} = \frac{b}{\sin 26}\]
\(b = \ \frac{\sin 26 \times 3.80}{\sin 62} = \ \frac{0.4384 \times 3.8}{0.8829} = 1.89cm\)
Find the length of the c side of the \(\mathrm{\Delta}ABC\) which have \(a = 5.3cm,\ b = 8.9cm\ \)and\(\ \angle C = 55{^\circ}\) using the cosine rule
5.9cm
7.3cm
6.7cm
8.8cm
Using the cosine rule, \(c^{2} = a^{2} + b^{2} - 2ab\cos C\)
\[c^{2} = {5.3}^{2} + {8.9}^{2} - 2(5.3)(8.9)\cos 55\]
\[c^{2} = 28.09 + 79.21 - 2(5.3)(8.9)(0.5736)\]
\[c^{2} = 28.09 + 79.21 - 54.11\]
\(c = \sqrt{53.19} = 7.3cm\)
Solve the \(\mathrm{\Delta}ABC\) which have \(\angle A = 145{^\circ},\ \angle B = 23{^\circ}\ \)and\(\ b = 5.02cm\) using the sine rule
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, \(\frac{b}{\sin B} = \frac{c}{\sin C}\)
\[\frac{5.02}{\sin 23} = \frac{c}{\sin 12}\]
\[\frac{5.02}{0.3907} = \frac{c}{0.2079}\]
\[c = \frac{5.02 \times 0.2079}{0.3907} \approx 2.67cm\]
Find the length of the c side of the \(\mathrm{\Delta}ABC\) which have \(a = 2.25cm,\ b = 6.24cm\ \)and\(\ \angle C = 78{^\circ}\) using the cosine rule
Using the cosine rule, \(c^{2} = a^{2} + b^{2} - 2ab\cos C\)
\[c^{2} = {2.25}^{2} + {6.24}^{2} - 2(2.25)(6.24)\cos 78\]
\[c^{2} = 5.0625 + 38.9376 - 2(2.25)(6.24)(0.2079)\]
\[c^{2} = 5.0625 + 38.9376 - 5.837832\]
\[c^{2} = 38.162268\]
\[c = \sqrt{38.162268} = 6.18cm\]
The angle of depression of a boat \(A\) from the top of a cliff \(21m\) high is \(42{^\circ}\). The angle of depression of another boat \(B\) in the straight line as \(A\) and from the same point is \(25{^\circ}\). Find the distance between the two boats.
\(distance\ bewteen\ boats\ A\ and\ B = distance\ of\ B\ from\ cliff - disatnce\ of\ A\ from\ cliff\)
\[disatnce\ of\ A\ from\ cliff:\]
\[\tan{42{^\circ}} = \frac{21}{OA}\]
\[OA = \frac{21}{\tan{42{^\circ}}} = \frac{21}{0.9004} = 23.32m\]
\[disatnce\ of\ B\ from\ cliff:\]
\[\tan{25{^\circ}} = \frac{21}{OB}\]
\[OB = \frac{21}{\tan{25{^\circ}}} = \frac{21}{0.4663} = 45.04m\]
\[distance\ bewteen\ boats\ A\ and\ B = distance\ of\ B\ from\ cliff - disatnce\ of\ A\ from\ cliff\]
\[distance\ bewteen\ boats\ A\ and\ B = 45.04 - 23.32 = \ 21.72m\]
Alice starts a \(3km\) walk from a point \(P\) on a bearing \(023{^\circ}\). She then walks \(4km\) on a bearing \(113{^\circ}\) to \(Q\). What is the distance of \(Q\) from \(P\)?
\(\angle O = 23{^\circ} + (180{^\circ} - 113{^\circ})\)
\[\angle O = 23{^\circ} + 67{^\circ}\]
\[\angle O = 90{^\circ}\]
Using the cosine rule,
\[o^{2} = p^{2} + q^{2} - 2pq\cos O\]
\[o^{2} = 4^{2} + 3^{2} - 2(4)(3)\cos 90\]
\[o^{2} = 16 + 9 - 0\]
\[o^{2} = 25\]
\[o = \sqrt{25} = \ 5km\]