Gambar rajah berikut menunjukkan sebuah segi tiga bersudut tegak.
a) panjang \(PR\)
\(\begin{aligned} PR&= \sqrt{15^2 + 8^2} \\\\&=\sqrt{289} \\\\&= 17 \text{ cm}. \end{aligned}\)
b) \(\text{sin }\angle PRQ\)
\(\begin{aligned} \text{sin}&=\dfrac{\text{bertentangan}}{\text{hipotenus}} \\\\&=\dfrac{PQ}{PR} \\\\&=\dfrac{15}{17}. \end{aligned}\)
c) \(\text{kos }\angle PRQ\)
\(\begin{aligned} \text{kos}&=\dfrac{\text{bersebelahan}}{\text{hipotenus}} \\\\&=\dfrac{QR}{PR} \\\\&=\dfrac{8}{17}. \end{aligned}\)
d) \(\text{tan }\angle QPR\)
\(\begin{aligned} \text{tan}&=\dfrac{\text{bertentangan}}{\text{bersebelahan}} \\\\&=\dfrac{QR}{PQ} \\\\&=\dfrac{8}{15}. \end{aligned}\)
Diberi \(\text{sin }\theta=0.6\) dan \(\text{kos }\theta=0.8\).
Berapakah nilai \(\text{tan }\theta\)?
Gambar rajah di bawah menunjukkan sebuah segi tiga bersudut tegak \(PQR\).
Diberi \(PR= 20 \text{ cm}\) dan \(\text{sin } \angle QPR= \dfrac{3}{5}\).
a) Tentukan panjang \(QR\).
Perlu difahami bahawa \(\text{sin } \angle QPR= \dfrac{3}{5}\).
Maka,
\(\begin{aligned} \text{sin } \angle QPR &=\dfrac{3}{5} \\\\ \dfrac{QR}{PR}&=\dfrac{3}{5} \\\\ \dfrac{QR}{20}&=\dfrac{3}{5} \\\\ QR&=\dfrac{3}{5}\times20 \\\\ QR&= 12 \text{ cm}. \end{aligned}\)
b) Hitung nilai \(\text{kos }\angle QPR\)
Pertama, kita perlu hitung panjang \(PQ\).
\(\begin{aligned} PQ&=\sqrt{PR^2-QR^2} \\\\&= \sqrt{20^2-12^2} \\\\&= \sqrt{256} \\\\&= 16 \text{ cm}. \\\\\end{aligned}\)
\(\begin{aligned} \text{kos } \angle QPR&=\dfrac{PQ}{PR} \\\\&=\dfrac{16}{20} \\\\&=\dfrac{4}{5}. \end{aligned}\)
Perlu difahami bahawa,
\(\begin{aligned}1^\circ&=60' \\\\1'&=60'' \end{aligned}\)
Uji diri anda dengan latihan bertahap
Ada yang tidak kena dengan soalan ini.