5.2.3 Nisbah Trigonometri, Praktis Berformat PT3

Soalan 7:
Dalam rajah di bawah, AEC dan BCD ialah garis lurus. E ialah titik tengah AC.


$\text{Diberi kos }x=\frac{5}{13}\text{ dan }\sin y=\frac{3}{5}$
(a) cari nilai tan x.
(b) Hitung panjang, dalam cm, BD.

Penyelesaian:
$\begin{array}{l}\text{(a)}\\ \text{Diberi kos }x=\frac{5}{13},\text{ oleh itu }BC=5,AB=13\\ AC=\sqrt{{13}^2-5^2}\\ =\sqrt{169-25}\\ =\sqrt{144}\\ =12\text{ cm}\\ \\ \text{tan }x=\frac{AC}{BC}=\frac{12}{5}\end{array}$

$\begin{array}{l}\text{(b)}\\ \text{Bagi }\Delta DCE:\\ \sin y=\frac{3}{5}\\ \frac{EC}{DE}=\frac{3}{5}\\ \frac{EC}{10}=\frac{3}{5}\\ EC=\frac{3}{5}\times 10=6\text{ cm}\\ \\ D{C}^2={10}^2-6^2\\ =64\\ DC=8\text{ cm}\\ \\ \text{Bagi }\Delta ABC:\\ AC=2\times 6=12\text{ cm}\\ \\ \tan x=\frac{12}{5}\\ \frac{12}{CB}=\frac{12}{5}\\ CB=5\text{ cm}\\ \\ BD=DC+CB\\ =\text{8 cm + 5 cm}\\ =\text{13 cm}\end{array}$

Soalan 8:
Dalam rajah di bawah, T ialah titik tengah garis PR.


(a) Cari nilai tan x^o.
(b) Hitung panjang, dalam cm, PQ.

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

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