# 9.3.1 Trigonometry II, SPM Paper 1 (Short Questions)

Question 1:

In the diagram above, find the value of tan θ.

Solution:

$\begin{array}{l}\text{In}△\text{}ABC,\text{using Pythagoras’ Theorem,}\\ AC=\sqrt{{1}^{2}+{1}^{2}}=\sqrt{2}\text{}cm\\ \\ \mathrm{tan}\theta =\frac{CD}{AC}\\ \mathrm{tan}\theta =\frac{1}{\sqrt{2}}\end{array}$

Question 2:

In the diagram above, ABCE is a rectangle and point D lies on the straight line EC. Given that DC = 5 cm and AE = 4cm, find the value of cosθ.

Solution:
$\begin{array}{l}AD=DC=5cm\\ \\ \text{In}△\text{}AED,\text{using Pythagoras’ Theorem,}\\ ED=\sqrt{{5}^{2}-{4}^{2}}=3cm\\ \\ \mathrm{cos}\theta =-\mathrm{cos}\angle ADE←\overline{)\begin{array}{l}\text{Since}{90}^{\circ }<\theta <{180}^{\circ }\\ \left(\text{in quadrant II),}\mathrm{cos}\theta \text{is negative}\end{array}}\\ \mathrm{cos}\theta =-\frac{ED}{AD}\\ \mathrm{cos}\theta =-\frac{3}{5}\end{array}$

Question 3:

In the diagram above, PMR is a straight line, M is the midpoint of line PR. Given that QR = 12cm and sin y°= 0.6, find the value of tan x°.

Solution:
$\begin{array}{l}\text{In triangle}QMR\text{,}\\ \mathrm{sin}{y}^{\circ }=0.6\\ \mathrm{sin}{y}^{\circ }=\frac{QR}{QM}=\frac{6}{10}\\ \text{Given}QR=12cm,\text{}\therefore QM=10×2=20cm\\ \\ \text{In}△\text{}QMR,\text{using Pythagoras’ Theorem,}\\ MR=\sqrt{{20}^{2}-{12}^{2}}=16cm\\ PR=16×2=32cm\\ \\ \text{Hence}\mathrm{tan}{x}^{\circ }=\frac{QR}{PR}=\frac{12}{32}=\frac{3}{8}\end{array}$