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}

, calculate the length, in cm, of AB.

Solution:

\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{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} .

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

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

Solution:

(a)

\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{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}