5.2.1 Trigonometric Ratios, PT3 Focus Practice - Mathematics

5.2.1 Trigonometric Ratios, PT3 Focus Practice

Question 1:

Diagram below shows a right-angled triangle ABC.

It is given that

cos x^{o} = \frac{5}{13}

$\cos x^{o} = \frac{5}{13}$ , calculate the length, in cm, of AB.

Solution:

\begin{matrix} cos x^{o} = \frac{A B}{A C} cos x^{o} = \frac{5}{13} \frac{A B}{39} = \frac{5}{13} A B = \frac{5}{13} \times 39 = 15 cm \end{matrix}

$\begin{array}{l} \cos x^{o} = \frac{A B}{A C} \\ \cos x^{o} = \frac{5}{13} \\ \frac{A B}{39} = \frac{5}{13} \\ A B = \frac{5}{13} \times 39 \\ = 15 cm \end{array}$

Question 2:

Diagram below consists of two right-angled triangles.

Determine the value of cos x^o.

Solution:

\begin{matrix} A C = \sqrt{13^{2} - 12^{2}} = \sqrt{25} = 5 cm C D = \sqrt{5^{2} - 3^{2}} = \sqrt{16} = 4 cm cos x^{o} = \frac{C D}{A C} = \frac{4}{5} \end{matrix}

$\begin{array}{l} A C = \sqrt{13^{2} - 12^{2}} \\ = \sqrt{25} \\ = 5 cm \\ C D = \sqrt{5^{2} - 3^{2}} \\ = \sqrt{16} \\ = 4 cm \\ \cos x^{o} = \frac{C D}{A C} \\ = \frac{4}{5} \end{array}$

Question 3:

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

ABC and EBD are straight lines.

It is given that

s i n x^{o} = \frac{5}{13} and cos y^{o} = \frac{3}{5} .

$s i n x^{o} = \frac{5}{13} and \cos y^{o} = \frac{3}{5} .$

(a) Find the value of tan x^o.

(b) Calculate the length, in cm, of ABC.

Solution:

(a)

\begin{matrix} s i n x^{o} = \frac{5}{13}, D C = 13 cm B C = \sqrt{13^{2} - 5^{2}} = \sqrt{144} = 12 cm Thus, tan x^{o} = \frac{5}{12} \end{matrix}

$\begin{array}{l} s i n x^{o} = \frac{5}{13}, D C = 13 cm \\ B C = \sqrt{13^{2} - 5^{2}} \\ = \sqrt{144} \\ = 12 cm \\ Thus, \tan x^{o} = \frac{5}{12} \end{array}$

(b)

\begin{matrix} cos y^{o} = \frac{A B}{15} \frac{3}{5} = \frac{A B}{15} A B = 9 cm Thus, A B C = 9 + 12 = 21 cm \end{matrix}

$\begin{array}{l} \cos y^{o} = \frac{A B}{15} \\ \frac{3}{5} = \frac{A B}{15} \\ A B = 9 cm \\ Thus, A B C = 9 + 12 \\ = 21 cm \end{array}$

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