**15.2.1 Trigonometry, PT3 Focus Practice**

**Question 1:**

Diagram below shows a right-angled triangle

*ABC*.It is given that $\mathrm{cos}{x}^{o}=\frac{5}{13}$
, calculate the length, in cm, of

*AB*.

Solution:Solution:

**Question 2:**

In the diagram,

*PQR*and*QTS*are straight lines.It is given that
$\mathrm{tan}y=\frac{3}{4}$
, calculate the length, in cm, of

*RS*.

Solution:Solution:

**Question 3:**

In the diagram,

*PQR*is a straight line.It is given that
$\mathrm{cos}{x}^{o}=\frac{3}{5}$
, hence sin

*y*^{o}=

*Solution:*
$\begin{array}{l}\mathrm{cos}{x}^{o}=\frac{PQ}{PS}\\ \frac{PQ}{10}=\frac{3}{5}\\ PQ=\frac{3}{5}\times 10\\ \text{}=6\text{cm}\\ QR=PR-PQ\\ =21-6\\ =15\text{cm}\end{array}$

$\begin{array}{l}Q{S}^{2}={10}^{2}-{6}^{2}\leftarrow \overline{)\text{pythagoras'Theorem}}\\ \text{}=100-36\\ \text{}=64\\ QS=\sqrt{64}\\ \text{}=8\text{cm}\\ R{S}^{2}={15}^{2}+{8}^{2}\\ \text{}=225+64\\ \text{}=289\\ RS=\sqrt{289}\\ \text{}=17\text{cm}\\ \mathrm{sin}{y}^{o}=\frac{15}{17}\end{array}$

**Question 4:**

Diagram below consists of two right-angled triangles.

Determine the value of cos

*x*^{o}.

Solution:Solution:

**Question 5:**

Diagram below consists of two right-angled triangles

*ABE*and*DBC*.*ABC*and

*EBD*are straight lines.

It is given that
$sin{x}^{o}=\frac{5}{13}\text{and}\mathrm{cos}{y}^{o}=\frac{3}{5}.$

(a) Find the value of tan

*x*^{o}.(b) Calculate the length, in cm, of

*ABC.*

Solution:Solution:

**(a)**

$\begin{array}{l}sin{x}^{o}=\frac{5}{13},\text{}DC=13\text{cm}\\ BC=\sqrt{{13}^{2}-{5}^{2}}\\ \text{}=\sqrt{144}\\ \text{}=12\text{cm}\\ \text{Thus},\text{}\mathrm{tan}{x}^{o}=\frac{5}{12}\end{array}$

**(b)**

$\begin{array}{l}\mathrm{cos}{y}^{o}=\frac{AB}{15}\\ \text{}\frac{3}{5}=\frac{AB}{15}\\ \text{}AB=9\text{cm}\\ \text{Thus,}ABC=9+12\\ \text{}=21\text{cm}\end{array}$