Lanjutan Contoh Soal Bab Trigonometri (2) (K13 Revisi)

\begin{array}{ll}\\ \fbox{11}.&\textrm{Jika}\: \: x_{1}\: \: \textrm{dan}\: \: x_{2}\: \: \textrm{memenuhi}\: \: 12\cos ^{2}x-\cos x-1=0.\\ &\textrm{maka nilai}\: \: \sec ^{2}x_{1}+\sec ^{2}x_{2}=....\\ &\begin{array}{llllll}\\ \textrm{a}.&22&&&\textrm{d}.&25\\ \textrm{b}.&23&\textrm{c}.&24&\textrm{e}.&26\\ \end{array}\\ &\\ &\textrm{Jawab}:\quad\textbf{d}\\ &\begin{aligned}12\cos ^{2}x+\cos x-1&=0\\ \left ( 3\cos x-1 \right )\left ( 4\cos x+1 \right )&=0\\ \cos x_{1}=\displaystyle \frac{1}{3}\quad \textrm{atau}\quad \cos x_{2}&=-\displaystyle \frac{1}{4}\\ &\begin{cases} \sec x_{1} & =\displaystyle \frac{1}{\cos x_{1}} \\ \Leftrightarrow &\sec ^{2}x_{1}=\displaystyle \frac{1}{\cos^{2} x_{1}}=3^{2}=9\\ \sec x_{2} & =\displaystyle \frac{1}{\cos x_{2}}\\ \Leftrightarrow &\sec ^{2}x_{2}=\displaystyle \frac{1}{\cos^{2} x_{2}}=(-4)^{2}=16 \end{cases}\\ \sec ^{2}x_{1}+\sec ^{2}x_{2}&=9+16=25 \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{12}.&\textrm{Jika}\: \: \displaystyle \frac{6\sin \gamma \cos \gamma }{2\sin ^{2}\gamma }=\displaystyle \frac{1}{2}\: \: \textrm{dengan}\: \: 0^{\circ}<\gamma <180^{\circ},\\ &\textrm{maka nilai}\: \: \tan \gamma =....\\ &\begin{array}{llllll}\\ \textrm{a}.&1&&&\textrm{d}.&6\\ \textrm{b}.&2&\textrm{c}.&4&\textrm{e}.&8\\\\ &&&&&(\textbf{SAT Subject Test}) \end{array}\\ &\\ &\textrm{Jawab}:\quad\textbf{d}\\ &\begin{aligned}\displaystyle \frac{6\sin \gamma \cos \gamma }{2\sin ^{2}\gamma }&=\displaystyle \frac{1}{2}\\ \displaystyle \frac{6\sin \gamma \cos \gamma }{2\sin \gamma \sin \gamma }&=\displaystyle \frac{1}{2}\\ \displaystyle \frac{\cos \gamma }{\sin \gamma }&=\displaystyle \frac{1}{6}\\ \cot \gamma &=\displaystyle \frac{1}{6},\: \textrm{maka}\\ \therefore \: \: \: \tan \gamma &=6 \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{13}.&\textrm{Jika}\: \: \sin \theta =p,\: \: \textrm{maka}\: \: \displaystyle \frac{1}{2p^{2}-3+\displaystyle \frac{1}{p^{2}}}=....\\ &\begin{array}{llllll}\\ \textrm{a}.&\sec ^{2}\theta \tan ^{2}\theta &&&\textrm{d}.&\displaystyle \frac{\tan ^{2}\theta }{2\cos ^{2}\theta -1}\\ \textrm{b}.&\sec ^{2}\theta \cos \theta &\textrm{c}.&\left ( 2\cos ^{2}-1 \right )\sec \theta &\textrm{e}.&\displaystyle \frac{1+\cot \theta }{\sin \theta \cos \theta }\\ \end{array}\\ &\\ &\textrm{Jawab}:\quad\textbf{d}\\ &\begin{aligned}\displaystyle \frac{1}{2p^{2}-3+\displaystyle \frac{1}{p^{2}}}&=\displaystyle \frac{1}{\left ( p-\displaystyle \frac{1}{p} \right )\left ( 2p-\displaystyle \frac{1}{p} \right )}\\ &=\displaystyle \frac{p^{2}}{\left ( p^{2}-1 \right )\left ( 2p^{2}-1 \right )}\\ &=\displaystyle \frac{\sin ^{2}\theta }{\left ( \sin ^{2}\theta -1 \right )\left ( 2\sin ^{2}\theta -1 \right )}\\ &=\displaystyle \frac{\sin ^{2}\theta }{\left ( -\cos ^{2}\theta \right )\left ( 2\left ( 1-\cos ^{2}\theta \right ) -1 \right )}\\ &=\displaystyle \frac{\sin ^{2}\theta }{\cos ^{2}\theta }\times \displaystyle \frac{1}{2\cos ^{2}\theta -1}\\ &=\displaystyle \frac{\tan ^{2}\theta }{2\cos ^{2}\theta -1} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{14}.&\textrm{Jika diketahui}\: \: \displaystyle \frac{\cos ^{2}\alpha -\sin ^{2}\alpha }{\sin \alpha \cos \alpha }=p,\: \: \textrm{maka}\: \: \cot ^{2}\alpha +\tan ^{2}\alpha =....\\ &\begin{array}{llllll}\\ \textrm{a}.&p^{2}+2&&&\textrm{d}.&1-p^{2}\\ \textrm{b}.&p^{2}+1&\textrm{c}.&p^{2}&\textrm{e}.&2-p^{2}\\\\ &&&&&(\textbf{SIMAK UI 2013 Mat Das}) \end{array}\\ &\\ &\textrm{Jawab}:\quad\textbf{a}\\ &\begin{aligned}\displaystyle \frac{\cos ^{2}\alpha -\sin ^{2}\alpha }{\sin \alpha \cos \alpha }&=p\\ \displaystyle \frac{\cos \alpha }{\sin \alpha }-\displaystyle \frac{\sin \alpha }{\cos \alpha }&=p\\ \cot \alpha -\tan \alpha &=p\qquad (\textrm{dikuadratkan})\\ \cot ^{2}\alpha -2+\tan ^{2}\alpha &=p^{2}\\ \cot ^{2}\alpha +\tan ^{2}\alpha &=p^{2}+2 \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{15}.&\textrm{Jika diketahui}\: \: \sin \beta -\tan \beta -2\cos \beta +2=0\: \: \textrm{dengan}\: \: 0<\beta <\displaystyle \frac{\pi }{2},\\ &\textrm{maka himpunan harga}\: \: \sin \beta =....\\ &\begin{array}{llllll}\\ \textrm{a}.&\left \{ \displaystyle \frac{2}{5}\sqrt{5} \right \}&&&\textrm{d}.&\left \{ \displaystyle \frac{1}{5}\sqrt{5} \right \}\\ \textrm{b}.&\left \{ 0 \right \}&\textrm{c}.&\left \{ \displaystyle \frac{2}{5}\sqrt{5},0 \right \}&\textrm{e}.&\left \{ \displaystyle \frac{1}{5}\sqrt{5},0 \right \}\\\\ &&&&&(\textbf{SIMAK UI 2009}) \end{array}\\ &\\ &\textrm{Jawab}:\quad\textbf{a}\\ &\begin{aligned}\sin \beta -\tan \beta -2\cos \beta +2&=0\\ \sin \beta -\displaystyle \frac{\sin \beta }{\cos \beta } -2\cos \beta +2&=0\\ \sin \beta \cos \beta -\sin \beta -2\cos^{2} \beta +2\cos \beta &=0\\ \sin \beta \left ( \cos \beta -1 \right )-2\cos \beta \left ( \cos \beta -1 \right )&=0\\ \left ( \sin \beta -2\cos \beta \right )\left ( \cos \beta -1 \right )&=0\\ \left ( \sin \beta -2\cos \beta \right )=0\: (\textbf{mm})\quad \textrm{atau}\quad \left ( \cos \beta -1 \right )&=0\: (\textbf{tmm})\\ \textrm{maka},\qquad\qquad\quad&\\ \left ( \sin \beta -2\cos \beta \right )&=0\\ \sin \beta &=2\cos \beta\\ \displaystyle \frac{\sin \beta }{\cos \beta }&=2\\ \tan \beta &=2=\displaystyle \frac{2}{1},\qquad \textrm{buatlah ilustrasi} \\ &\textrm{dengan membuat segitiga siku-siku}.\\ \textrm{Sehingga akan didapatkan nilai}&\\ \sin \beta &=\displaystyle \frac{2}{5}\sqrt{5} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{16}.&\textrm{Diketahui panjang dua sisi sebuah segitiga adalah 10 cm dan 8 cm}.\\ &\textrm{Nilai keliling dari segitiga tersebut yang tidak mungkin adalah}\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&\textrm{32 cm}&&&\textrm{d}.&\textrm{35 cm}\\ \textrm{b}.&\textrm{33 cm}&\textrm{c}.&\textrm{34 cm}&\textrm{e}.&\textrm{36 cm}\\\\ &&&&&(\textbf{SNMPTN 2010 Mat IPA}) \end{array}\\ &\\ &\textrm{Jawab}:\quad\textbf{e}\\ &\begin{array}{|c|c|c|}\hline \multicolumn{3}{|c|}{\begin{aligned}&\\ \textrm{Misalkan sisi-sisi}&\textrm{nya}\\ &\left\{\begin{matrix} a=10\\ b=8\\ c=c \end{matrix}\right.\\ \textrm{maka}&\quad a+b>c\: (\textrm{definisi})\\ &\quad 10+8>c \Leftrightarrow 18>c\\ &\end{aligned}}\\\hline \textrm{No}.&\textrm{Keliling(K)},\: K=(a+b)+\textbf{c}&\textrm{Keterangan}\\\hline \textrm{a}&32=18+(14)&\textrm{Boleh}\\\hline \textrm{b}&33=18+(15)&\textrm{Boleh}\\\hline \textrm{c}&34=18+(16)&\textrm{Boleh}\\\hline \textrm{d}&35=18+(17)&\textrm{Boleh}\\\hline \textbf{e}&36=18+(18)&\textbf{Tidak boleh}\\\hline \end{array} \end{array}.

\begin{array}{ll}\\ \fbox{17}.&\textrm{Jika dalam segitiga PQR diketahui bahwa QR = 10, PR = 40}\\ &\textrm{dan}\: \: \angle R=120^{\circ},\: \textrm{maka PQ =}\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&10\sqrt{13}&&&\textrm{d}.&10\sqrt{17}\\ \textrm{b}.&20\sqrt{13}&\textrm{c}.&10\sqrt{21}&\textrm{e}.&\sqrt{5}\\ \end{array}\\ &\\ &\textrm{Jawab}:\quad\textbf{c}\\ &\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\begin{aligned}\textrm{Diketahui bahwa}&:\\ &\begin{cases} QR &=10 \\ PR &=40 \\ \angle R &=120^{\circ} \end{cases}\\ & \end{aligned}}\\\hline \textrm{Gambar}&\textrm{Solusi}\\\hline \begin{aligned}&\textrm{perhatikanlah}\\ &\textrm{ilustrasi gambar}\\ &\textrm{di bawah} \end{aligned}&\begin{aligned}\textrm{Dengan}\: \, &\textbf{aturan cosinus}\\ PR^{2}&=PR^{2}+QR^{2}-2.PR.QR.\cos \angle R\\ &=40^{2}+10^{2}-2.40.10.\cos 120^{\circ}\\ &=1600+100-800.\left ( -\displaystyle \frac{1}{2} \right )\\ &=1700+400\\ PR^{2}&=2100\\ PR&=\sqrt{2100}\\ &=10\sqrt{21} \end{aligned}\\\hline \end{array} \end{array}..

\begin{array}{ll}\\ \fbox{18}.&\textrm{Jika dalam segitiga ABC diketahui bahwa}\: \: \angle A=60^{\circ},\: \: \angle B=75^{\circ}\\ &\textrm{dan}\: \: \textrm{BC = 3},\: \textrm{maka panjang sisi AB =}\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&3\sqrt{2}&&&\textrm{d}.&6\sqrt{2}\\ \textrm{b}.&3\sqrt{6}&\textrm{c}.&\sqrt{6}&\textrm{e}.&3\sqrt{3}\\ \end{array}\\ &\\ &\textrm{Jawab}:\quad\textbf{c}\\ &\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\begin{aligned}\textrm{Diketahui bahwa}&:\: \textrm{pada}\: \: \triangle \textrm{ABC}\\ &\begin{cases} \angle A &=60^{\circ} \\ \angle B&=75^{\circ} \\ \angle C&=180^{\circ}-\left ( 60^{\circ}+75^{\circ} \right )=45^{\circ}\\ BC&=3 \end{cases}\\ & \end{aligned}}\\\hline \textrm{Gambar}&\textrm{Solusi}\\\hline \begin{aligned}&\textrm{perhatikanlah}\\ &\textrm{ilustrasi gambar}\\ &\textrm{di bawah} \end{aligned}&\begin{aligned}\textrm{Dengan}\: \, &\textbf{aturan sinus},\: \textrm{yaitu}:\\ \displaystyle \frac{AB}{\sin \angle C}&=\displaystyle \frac{BC}{\sin \angle A}\\ \displaystyle \frac{AB}{\sin 45^{\circ}}& =\displaystyle \frac{3}{\sin 60^{\circ}}\\ AB&=\displaystyle \frac{3}{\sin 60^{\circ}}\times \sin 45^{\circ}\\ &=\displaystyle \frac{3}{\displaystyle \frac{1}{2}\sqrt{3}}\times \displaystyle \frac{1}{2}\sqrt{2}\\ &=\sqrt{6} \end{aligned}\\\hline \end{array} \end{array}..

\begin{array}{ll}\\ \fbox{19}.&\textrm{Panjang sisi AD sebuah segiempat ABCD berikut adalah}... \end{array}..

\begin{array}{ll}\\ .\, \: \: \quad&\begin{array}{llllll}\\ \textrm{a}.&2\sqrt{7}\: \: \textrm{cm}&&&\textrm{d}.&8\: \: \textrm{cm}\\ \textrm{b}.&4\sqrt{6}\: \: \textrm{cm}&\textrm{c}.&2\sqrt{19}\: \: \textrm{cm}&\textrm{e}.&6\: \: \textrm{cm}\\\\ &&&&&(\textbf{Soal UN 2015}) \end{array}\\ &\\ &\textrm{Jawab}:\quad\textbf{a}\\ &\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\textrm{Proses}}\\\hline \textrm{Aturan Sinus}&\textrm{Aturan Cosinus}\\\hline \multicolumn{2}{|c|}{\textrm{saat mencari}}\\\hline \textrm{AC}&\textrm{AD}\\\hline \begin{aligned}\displaystyle \frac{AC}{\sin \angle B}&=\displaystyle \frac{BC}{\sin \angle BAC}\\ \displaystyle \frac{AC}{\sin 45^{\circ}}&=\displaystyle \frac{5\sqrt{2}}{\sin 30^{\circ}}\\ AC&=\displaystyle \frac{5\sqrt{2}}{\sin 30^{\circ}}\times \sin 45^{\circ}\\ &=\displaystyle \frac{5\sqrt{2}}{\displaystyle \frac{1}{2}}\times \displaystyle \frac{1}{2}\sqrt{2}\\ &=10 \end{aligned}&\begin{aligned}AD^{2}&=AC^{2}+CD^{2}-2.AC.CD.\cos 30^{\circ}\\ &=10^{2}+\left ( 4\sqrt{3} \right )^{2}-2.10.\left (4\sqrt{3} \right ).\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )\\ &=100+48-120\\ AD^{2}&=28\\ AD&=\sqrt{48}\\ &=\sqrt{4.7}\\ &=2\sqrt{7} \end{aligned}\\\hline \end{array} \end{array}..

\begin{array}{ll}\\ \fbox{20}.&\textrm{Jika pada segitiga ABC lancip dengan AB}=2\sqrt{2},\: BC=2,\: \textrm{dan}\\ &\angle ABC=\theta ,\: \sin \theta =\displaystyle \frac{1}{3},\: \textrm{maka AC}=....\\ &\begin{array}{llllll}\\ \textrm{a}.&\displaystyle \frac{1}{3}\sqrt{2}&&&\textrm{d}.&\displaystyle \frac{3}{2}\sqrt{2}\\ \textrm{b}.&\sqrt{6}&\textrm{c}.&\displaystyle \frac{2}{3}\sqrt{3}&\textrm{e}.&\displaystyle \frac{1}{2}\sqrt{2}\\\\ &&&&&(\textbf{UM UGM 2010 Mat Das}) \end{array}\\ &\\ &\textrm{Jawab}:\quad\textbf{c}\\ &\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\textrm{Proses penyelesaian}}\\\hline \textrm{Mencari nilai}&\textrm{Aturan Cosinus}\\\cline{2-2} \cos \theta &\textrm{AC}\\\hline \begin{aligned}\cos \theta &=\sqrt{1-\sin ^{2}\theta}\\ &=\sqrt{1-\left ( \displaystyle \frac{1}{3} \right )^{2}}\\ &=\sqrt{\displaystyle \frac{8}{9}}\\ &=\displaystyle \frac{2}{3}\sqrt{2}\\ & \end{aligned}&\begin{aligned}AC^{2}&=AB^{2}+BC^{2}-2.AB.BC.\cos \theta \\ &=\left ( 2\sqrt{2} \right )^{2}+2^{2}-2.\left ( 2\sqrt{2} \right ).2.\left ( \displaystyle \frac{2}{3}\sqrt{2} \right )\\ &=8+4-\displaystyle \frac{32}{3}\\ AC^{2}&=\displaystyle \frac{4}{3}\\ AC&=\displaystyle \frac{2}{3}\sqrt{3} \end{aligned}\\\hline \end{array} \end{array}..

 

Tentang ahmadthohir1089

Nama saya Ahmad Thohir asli orang Purwodadi, Jawa Tengah, lahir di Grobogan 02 Februari 1980. Pendidikan : Tingkat dasar lulus dari MI Nahdlatut Thullab di desa Manggarwetan,kecamatan Godong lulus tahun 1993. dan untuk tingkat menengah saya tempuh di MTs Nahdlatut Thullab Manggar Wetan lulus tahun 1996. Sedang untuk tingkat SMA saya menamatkannya di MA Futuhiyyah-2 Mranggen, Demak lulus tahun 1999. Setelah itu saya Kuliah di IKIP PGRI Semarang pada fakultas FPMIPA Pendidikan Matematika lulus tahun 2004. Pekerjaan : Sebagai guru (PNS DPK Kemenag) mapel matematika di MA Futuhiyah Jeketro, Gubug. Pengalaman mengajar : 1. GTT di MTs Miftahul Mubtadiin Tambakan Gubug tahun 2003 s/d 2005 2. GTT di SMK Negeri 3 Semarang 2005 s/d 2009 3. GT di MA Futuhiyah Jeketro Gubug sejak 1 September 2009
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