5.2.3 Trigonometric Ratios, PT3 Focus Practice - Mathematics

Question 7:
In diagram below, AEC and BCD are straight lines. E is the midpoint of AC.

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

$Given \cos x = \frac{5}{13} and \sin y = \frac{3}{5}$
(a) find the value of tan x.
(b) Calculate the length, in cm, of BD.

Solution:

\begin{matrix} (a) Given cos x = \frac{5}{13}, therefore B C = 5, A B = 13 A C = \sqrt{13^{2} - 5^{2}} = \sqrt{169 - 25} = \sqrt{144} = 12 cm tan x = \frac{A C}{B C} = \frac{12}{5} \end{matrix}

$\begin{array}{l} (a) \\ Given cos x = \frac{5}{13}, therefore B C = 5, A B = 13 \\ A C = \sqrt{13^{2} - 5^{2}} \\ = \sqrt{169 - 25} \\ = \sqrt{144} \\ = 12 cm \\ tan x = \frac{A C}{B C} = \frac{12}{5} \end{array}$

\begin{matrix} (b) For Δ D C E : sin y = \frac{3}{5} \frac{E C}{D E} = \frac{3}{5} \frac{E C}{10} = \frac{3}{5} E C = \frac{3}{5} \times 10 = 6 cm D C^{2} = 10^{2} - 6^{2} = 64 D C = 8 cm For Δ A B C : A C = 2 \times 6 = 12 cm tan x = \frac{12}{5} \frac{12}{C B} = \frac{12}{5} C B = 5 cm B D = D C + C B = 8 cm + 5 cm = 13 cm \end{matrix}

$\begin{array}{l} (b) \\ For Δ D C E : \\ \sin y = \frac{3}{5} \\ \frac{E C}{D E} = \frac{3}{5} \\ \frac{E C}{10} = \frac{3}{5} \\ E C = \frac{3}{5} \times 10 = 6 cm \\ D C^{2} = 10^{2} - 6^{2} \\ = 64 \\ D C = 8 cm \\ For Δ A B C : \\ A C = 2 \times 6 = 12 cm \\ \tan x = \frac{12}{5} \\ \frac{12}{C B} = \frac{12}{5} \\ C B = 5 cm \\ B D = D C + C B \\ = 8 cm + 5 cm \\ = 13 cm \end{array}$

Question 8:
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{matrix} (a) T R^{2} = 13^{2} - 12^{2} = 169 - 144 = 25 T R = \sqrt{25} = 5 cm tan x^{o} = \frac{12}{5} \end{matrix}

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

\begin{matrix} (b) P R = 2 \times 5 cm = 10 cm P Q^{2} = 10^{2} - 8^{2} = 100 - 64 = 36 P Q = \sqrt{36} = 6 cm \end{matrix}

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

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