5.3.1 Buku Teks (Cabaran Dinamis)

Uji Diri

Soalan 1:


Rajah menunjukkan segi tiga bersudut tegak KLM. Hitung
(a) θ dalam darjah dan minit
(b) sin (90^o – θ)
(c) kos (90^o – θ)

Penyelesaian:
(a)
$\begin{array}{l}\tan \theta =\frac{8}{15}\\ \theta ={\tan }^{-1}\left(\frac{8}{15}\right)\\ \theta ={28}^{o}4‘\end{array}$

(b)
$\begin{array}{l}\sin \left({90}^o-\theta \right)\\ =\sin \angle KML\\ =\frac{KL}{KM}\\ =\frac{15}{17}\leftarrow \boxed{KM=\sqrt{{15}^2+8^2}=17}\end{array}$

(c)
$\begin{array}{l}kos\left({90}^o-\theta \right)\\ =\text{kos}\angle KML\\ =\frac{LM}{KM}\\ =\frac{8}{17}\end{array}$

Soalan 2:


Rajah menunjukkan segi tiga bersudut tegak ABC. Diberi bahawa $\tan \theta =\frac{5}{12}.$ Hitung
(a) panjang AC, dalam cm
(b) nilai tan (90^o – θ)
(c) θ dalam darjah betul kepada tiga angka bererti

Penyelesaian:
(a)
$\begin{array}{l}\tan \theta =\frac{AB}{BC}\\ \frac{AB}{BC}=\frac{5}{12}\\ \frac{15}{BC}=\frac{5}{12}\\ BC=\frac{12}{5}\times 15\\ BC=36\\ \\ AC=\sqrt{{15}^2+{36}^2}\\ =\sqrt{1521}\\ =39\text{ cm}\end{array}$

(b)
$\begin{array}{l}tan\left({90}^o-\theta \right)\\ =\tan \angle CAB\\ =\frac{BC}{AB}\\ =\frac{36}{15}\\ =\frac{12}{5}\end{array}$

(c)
$\begin{array}{l}\tan \theta =\frac{15}{36}\\ \theta ={\tan }^{-1}\left(\frac{15}{36}\right)\\ \theta ={22.6}^o\end{array}$