6.3.2 Ratio and Graphs of Trigonometric Functions, SPM Paper (Short Questions)

Question 4:

In the diagram above, WZY is a straight line.

∠ X Y Z = 90^{o}, ∠ X W Z = 30^{o}

and WZ = XZ = 30 cm. Find the length of XY.

Solution:

\begin{array}{l} ∠ W X Z = ∠ X W Z = 30^{o} \\ ∴ ∠ X Z Y = 30^{o} + 30^{o} = 60^{o} \\ \sin ∠ X Z Y = \frac{X Y}{X Z} \\ \sin 60^{o} = \frac{X Y}{30} \\ X Y = \sin 60^{o} \times 30 \\ X Y = 25.98 c m \end{array}

Question 5:

In the diagram above, PQS is a right angle triangle. Given that SR = 6cm, PQ = 12 cm and 5SR = 2PS. Find the value of cos α and tan β.

Solution:

\begin{array}{l} 5 S R = 2 P S \\ P S = \frac{5}{2} S R \\ P S = \frac{5}{2} (6) \\ P S = 15 c m \\ \cos α = \frac{P Q}{P S} \\ \cos α = \frac{12}{15} = \frac{4}{5} \\ In △ P Q S, using Pythagoras’ Theorem, \\ Q S = \sqrt{P S^{2} - P Q^{2}} \\ Q S = \sqrt{15^{2} - 12^{2}} = 9 c m \\ \tan β = - \tan ∠ P S Q \leftarrow \begin{array}{l} Since 90^{\circ} < β < 180^{\circ} \\ (in quadrant II), tan β is negative \end{array} \\ \tan β = - \frac{P Q}{Q S} \\ \tan β = - \frac{12}{9} = - \frac{4}{3} \end{array}

Question 6:

In the diagram above, ADC is a straight line, if

\sin q = \frac{3}{5} and \tan p = \frac{1}{2}

. Find the distance of AC.

Solution:

\begin{array}{l} Given \sin q = \frac{B D}{A B} = \frac{3}{5} \\ \frac{B D}{30} = \frac{3}{5} \\ B D = \frac{3}{5} \times 30 \\ B D = 18 c m \\ In △ A B D, using Pythagoras’ Theorem, \\ A D = \sqrt{A B^{2} - B D^{2}} \\ A D = \sqrt{30^{2} - 18^{2}} = 24 c m \\ Given tan p = \frac{B D}{D C} = \frac{1}{2} \\ \frac{18}{D C} = \frac{1}{2} \\ D C = 36 c m \\ Hence, distance of A C = 24 + 36 = 60 c m . \end{array}