**Question 6:**

In diagram below,

*AEC*and

*BCD*are straight lines.

*E*is the midpoint of

*AC*.

$\text{Given}\mathrm{cos}x=\frac{5}{13}\text{and}\mathrm{sin}y=\frac{3}{5}$

(a) find the value of tan

*x.*

(b) Calculate the length, in cm, of

*BD*.

Solution:Solution:

$\begin{array}{l}\text{(a)}\\ \text{Givencos}x=\frac{5}{13},\text{therefore}BC=5,\text{}AB=13\\ AC=\sqrt{{13}^{2}-{5}^{2}}\\ \text{}=\sqrt{169-25}\\ \text{}=\sqrt{144}\\ \text{}=12\text{cm}\\ \\ \text{tan}x=\frac{AC}{BC}=\frac{12}{5}\end{array}$

$\begin{array}{l}\text{(b)}\\ \text{For}\Delta DCE:\\ \mathrm{sin}y=\frac{3}{5}\\ \frac{EC}{DE}=\frac{3}{5}\\ \frac{EC}{10}=\frac{3}{5}\\ EC=\frac{3}{5}\times 10=6\text{cm}\\ \\ D{C}^{2}={10}^{2}-{6}^{2}\\ \text{}=64\\ \text{}DC=8\text{cm}\\ \\ \text{For}\Delta ABC:\\ AC=2\times 6=12\text{cm}\\ \\ \mathrm{tan}x=\frac{12}{5}\\ \frac{12}{CB}=\frac{12}{5}\\ CB=5\text{cm}\\ \\ BD=DC+CB\\ =\text{8cm+5cm}\\ =\text{13cm}\end{array}$

**Question 7:**

In diagram below,

*T*is the midpoint of the line

*PR*.

(a) Find the value of tan

*x*

^{o}.

(b) Calculate the length, in cm, of

*PQ*.

*Solution:*$\begin{array}{l}\text{(a)}\\ T{R}^{2}={13}^{2}-{12}^{2}\\ \text{}=169-144\\ \text{}=25\\ TR=\sqrt{25}\\ \text{}=5\text{cm}\\ \mathrm{tan}{x}^{o}=\frac{12}{5}\end{array}$

$\begin{array}{l}\text{(b)}\\ PR=2\times 5\text{cm}\\ \text{}=10\text{cm}\\ P{Q}^{2}={10}^{2}-{8}^{2}\\ \text{}=100-64\\ \text{}=36\\ PQ=\sqrt{36}\\ \text{}=6\text{cm}\end{array}$

**Question 8:**

In diagram below,

*ABE*and

*DBC*are two right-angled triangles

*ABC*and

*DEB*are straight lines.

$\text{Itisgiventhat}\mathrm{cos}{y}^{o}=\frac{3}{5}.$

(a) Find the value of tan

*x*

^{o}.

(b) Calculate the length, in cm, of

*DE*.

*Solution:*$\text{(a)}\mathrm{tan}{x}^{o}=\frac{7}{24}$

$\begin{array}{l}\text{(b)}\\ \mathrm{cos}{y}^{o}=\frac{BC}{20}\\ \text{}\frac{3}{5}=\frac{BC}{20}\\ BC=\frac{3}{5}\times 20\\ \text{}=12\text{cm}\\ \\ \therefore B{D}^{2}={20}^{2}-{12}^{2}\\ \text{}=400-144\\ \text{}=256\\ BD=\sqrt{256}\\ \text{}=16\text{cm}\\ \\ DE=16-7\\ \text{}=9\text{cm}\end{array}$

**Question 9:**

Diagram below shows a vertical pole,

*PQ*. At 2.30 p.m. and 5.00 p.m., the shadow of the pole falls on

*QR*and

*QS*respectively.

Calculate

(a) the height, in m, of the pole.

(b) the value of

*w*.

*Solution:*

**(a)**

$\begin{array}{l}\text{tan}{55}^{o}=\frac{\text{Heightofthepole}}{3.2}\\ \text{Heightofthepole}=\text{tan}{55}^{o}\times 3.2\\ \text{}=1.428\times 3.2\\ \text{}=4.57\text{m}\end{array}$

**(b)**

**$\begin{array}{l}\text{tan}w=\frac{4.57}{3.20+2}\\ \text{}=\frac{4.57}{5.20}\\ \text{}=0.879\\ \text{}w={41}^{o}18\text{'}\end{array}$**