Solusi Soal Nomor 4 Uji Sertifikasi Pendidikan Menengah

$latex 4.\ \ \ (a)\ \ (i)\ \ {\bf Luas\ juring\ AOC}\\\\\begin{matrix} & & & \ \end{matrix}\begin{array}{rcl}\frac{Luas\ juring\ AOC}{luas\ lingkaran}&=&\frac{sudut\ pusat}{sudut\ lingkaran}\\\\Luas\ juring\ AOC&=&luas\ lingkaran\times \frac{\frac{\pi }{3}}{2\pi }\\\\&=&\pi \times r^2\times \frac{1}{6}\\\\&=&\pi \times 4^2\times \frac{1}{6}\\\\&=&\frac{8}{3}\pi \end{array}\\\\\begin{matrix} & & & \ \end{matrix}\qquad {\bf Luas\ persegi}=4\times 4=16\\\begin{matrix} & & & \ \end{matrix}\qquad Jadi\ {\bf luas\ daerah\ bangun\ OACMN}=\Big(16+\frac{8}{3}\pi \Big)\ m^2.$ $latex \begin{matrix} & & \quad \ \end{matrix}(ii)\ \ {\bf Panjang\ busur\ AC}\\\\\begin{matrix} & & & \ \end{matrix}\qquad \begin{array}{rcl}\frac{panjang\ busur\ AC}{keliling\ lingkaran}&=&\frac{sudut\ pusat}{sudut\ lingkaran}\\\\Panjang\ busur\ AC&=&keliling\ lingkaran\times \frac{\frac{\pi }{3}}{2\pi }\\\\&=&2\times \pi \times r\times \frac{1}{6}\\\\&=&2\times \pi \times 4\times \frac{1}{6}\\\\&=&\frac{4}{3}\pi \end{array}\\\\\\\begin{matrix} & & & \ \end{matrix}\qquad Jadi\ {\bf keliling\ bangun\ OACMN}\\\begin{matrix} & & & \ \end{matrix}\qquad =CM+MN+NO+OA+AC\\\begin{matrix} & & & \ \end{matrix}\qquad =4+4+4+4+\frac{4}{3}\pi \\\begin{matrix} & & & \ \end{matrix}\qquad =\Big(16+\frac{4}{3}\pi \Big)\ m$ $latex \begin{matrix} & \ \ \end{matrix}(b)\ \ Karena\ \frac{7\pi }{12}=\frac{\pi }{3}+\frac{\pi }{4}\ dan\ cos\ (\alpha +\beta )=cos\ \alpha cos\ \beta -sin\ \alpha sin\ \beta ,\\\\\begin{matrix} & & & \end{matrix}maka:\\\begin{matrix} & & & \ \end{matrix}\begin{array}{rcl}cos\ \frac{7\pi }{12}&=&cos\ \Big(\frac{\pi }{3}+\frac{\pi }{4}\Big)\\\\&=&cos\ \frac{\pi }{3}\ cos\ \frac{\pi }{4}-sin\ \frac{\pi }{3}\ sin\ \frac{\pi }{4}\\\\&=&\frac{1}{2}\times \frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{2}\times \frac{\sqrt{2}}{2}\\\\&=&\frac{\sqrt{2}-\sqrt{6}}{4}\end{array}$ $latex \begin{matrix} & \ \ \end{matrix}(c)\ \ Bukti:\\\\\begin{matrix} & & & \end{matrix}\begin{array}{rcl}\frac{1}{sec\ \theta +tan\ \theta }&=&\frac{1}{\frac{1}{cos\ \theta } +\frac{sin\ \theta }{cos\ \theta }}\\\\&=&\frac{cos\ \theta }{1+sin\ \theta }\qquad \quad \times \frac{1-sin\ \theta }{1-sin\ \theta }\\\\&=&\frac{cos\ \theta (1-sin\ \theta )}{1-sin^2\ \theta }\\\\&=&\frac{cos\ \theta (1-sin\ \theta )}{cos^2\ \theta }\\\\&=&\frac{1-sin\ \theta }{cos\ \theta }\end{array}$