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**15.2.1 Trigonometry, PT3 Focus Practice**

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**Question 1:**

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Diagram below shows a right-angled triangle *ABC*.

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

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Solution:

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$\begin{array}{l}\mathrm{cos}{x}^{o}=\frac{AB}{AC}\\ \mathrm{cos}{x}^{o}=\frac{5}{13}\\ \frac{AB}{39}=\frac{5}{13}\\ AB=\frac{5}{13}\times 39\\ \text{}=15\text{cm}\end{array}$
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**Question 2:**

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In the diagram, *PQR *and *QTS* are straight lines.

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It is given that
$\mathrm{tan}y=\frac{3}{4}$
, calculate the length, in cm, of *RS*.

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Solution:

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$\begin{array}{l}\text{In}\u25b3\text{}PQT,\\ \mathrm{tan}y=\frac{PQ}{QT}\\ \frac{3}{4}=\frac{6}{QT}\\ QT=6\times \frac{4}{3}\\ \text{}=8\text{cm}\\ \\ \text{In}\u25b3QRS,\text{}QS=8+8=16\text{cm}\\ \\ \therefore R{S}^{2}={12}^{2}+{16}^{2}\leftarrow \overline{)\text{pythagoras'Theorem}}\\ \text{}=144+256\\ \text{}=400\\ RS=\sqrt{400}\\ \text{}=20\text{cm}\end{array}$
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**Question 3:**

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In the diagram, *PQR *is a straight line.

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It is given that
$\mathrm{cos}{x}^{o}=\frac{3}{5}$
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, hence sin *y*^{o} =
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*Solution:*

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$\begin{array}{l}\mathrm{cos}{x}^{o}=\frac{PQ}{PS}\\ \frac{PQ}{10}=\frac{3}{5}\\ PQ=\frac{3}{5}\times 10\\ \text{}=6\text{cm}\\ QR=PR-PQ\\ =21-6\\ =15\text{cm}\end{array}$
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$\begin{array}{l}Q{S}^{2}={10}^{2}-{6}^{2}\leftarrow \overline{)\text{pythagoras'Theorem}}\\ \text{}=100-36\\ \text{}=64\\ QS=\sqrt{64}\\ \text{}=8\text{cm}\\ R{S}^{2}={15}^{2}+{8}^{2}\\ \text{}=225+64\\ \text{}=289\\ RS=\sqrt{289}\\ \text{}=17\text{cm}\\ \mathrm{sin}{y}^{o}=\frac{15}{17}\end{array}$
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**Question 4:**

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Diagram below consists of two right-angled triangles.

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Determine the value of cos *x*^{o}.

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Solution:

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$\begin{array}{l}AC=\sqrt{{13}^{2}-{12}^{2}}\\ \text{}=\sqrt{25}\\ \text{}=5\text{cm}\\ CD=\sqrt{{5}^{2}-{3}^{2}}\\ \text{}=\sqrt{16}\\ \text{}=4\text{cm}\\ \mathrm{cos}{x}^{o}=\frac{CD}{AC}\\ \text{}=\frac{4}{5}\end{array}$
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**Question 5:**

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Diagram below consists of two right-angled triangles *ABE* and *DBC*.

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*ABC *and *EBD* are straight lines.

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

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(a) ` `

Find the value of tan *x*^{o}.

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(b)

Calculate the length, in cm, of *ABC.*

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Solution:

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**(a)**

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

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