15.2.2 Trigonometry, PT3 Focus Practice

Posted on July 10, 2018 by user

Question 6:
In diagram below, AEC and BCD are straight lines. E is the midpoint of AC.

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

Solution:

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

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

Question 7:
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:

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

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

Question 8:
In diagram below, ABE and DBC are two right-angled triangles ABC and DEB are straight lines.

$It is given that \cos y^{o} = \frac{3}{5} .$
(a) Find the value of tan x^o.
(b) Calculate the length, in cm, of DE.

Solution:

$(a) \tan x^{o} = \frac{7}{24}$

$\begin{array}{l} (b) \\ \cos y^{o} = \frac{B C}{20} \\ \frac{3}{5} = \frac{B C}{20} \\ B C = \frac{3}{5} \times 20 \\ = 12 cm \\ ∴ B D^{2} = 20^{2} - 12^{2} \\ = 400 - 144 \\ = 256 \\ B D = \sqrt{256} \\ = 16 cm \\ D E = 16 - 7 \\ = 9 cm \end{array}$

Question 9:
Diagram below shows a vertical pole, PQ. At 2.30 p.m. and 5.00 p.m., the shadow of the pole falls on QR and QS respectively.

Calculate
(a) the height, in m, of the pole.
(b) the value of w.

Solution:
(a)

$\begin{array}{l} tan 55^{o} = \frac{Height of the pole}{3.2} \\ Height of the pole = tan 55^{o} \times 3.2 \\ = 1.428 \times 3.2 \\ = 4.57 m \end{array}$

(b)

$\begin{array}{l} tan w = \frac{4.57}{3.20 + 2} \\ = \frac{4.57}{5.20} \\ = 0.879 \\ w = 41^{o} 18' \end{array}$