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Tolong bantuannya teman teman​

Berikut ini adalah pertanyaan dari thorfin42 pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

Tolong bantuannya teman teman​
Tolong bantuannya teman teman​

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

\purple{\huge{1a.}}

Perhatikan gambar pertama (terlampir) :

Dengan rumus Pythagoras :

\begin{array}{lll}\text{QR}^2&=&\text{PQ}^2+\text{PR}^2\\~\\\text{PQ}^2&=&\text{QR}^2-\text{PR}^2\\~\\\text{PQ}&=&\sqrt{\text{QR}^2-\text{PR}^2}\\~\\&=&\sqrt{13^2-5^2}\\~\\&=&\sqrt{169-25}\\~\\&=&\sqrt{144}\\~\\&=&12\end{array}

\sin~a=\frac{\text{PQ}}{\text{QR}}~\to \red{\huge{\sin~a=\frac{12}{13}}}

\cos~a=\frac{\text{PR}}{\text{QR}}~\to \red{\huge{\cos~a=\frac{5}{13}}}

\tan~a=\frac{\text{PQ}}{\text{PR}}~\to \red{\huge{\tan~a=\frac{12}{5}}}

\\

\purple{\huge{1b.}}

Perhatikan gambar kedua (terlampir) :

Dengan rumus Pythagoras :

\begin{array}{lll}\text{LM}^2&=&\text{KL}^2+\text{KM}^2\\~\\\text{LM}&=&\sqrt{\text{KL}^2+\text{KM}^2}\\~\\&=&\sqrt{7^2+2^2}\\~\\&=&\sqrt{49+4}\\~\\&=&\sqrt{53}\end{array}

\sin~a=\frac{\text{KL}}{\text{LM}}~\to \red{\huge{\sin~a=\frac{7}{\sqrt{53}}=\frac{7}{53}\sqrt{53}}}

\cos~a=\frac{\text{KM}}{\text{LM}}~\to \red{\huge{\cos~a=\frac{2}{\sqrt{53}}=\frac{2}{53}\sqrt{53}}}

\tan~a=\frac{\text{KL}}{\text{KM}}~\to \red{\huge{\tan~a=\frac{7}{2}}}

\\

\purple{\huge{2a.}}

Perhatikan gambar ketiga (terlampir) :

Dengan rumus Pythagoras :

\begin{array}{lll}\text{OP}^2&=&\text{OA}^2+\text{AP}^2\\~\\\text{OP}&=&\sqrt{\text{OA}^2+\text{AP}^2}\\~\\&=&\sqrt{3^2+4^2}\\~\\&=&\sqrt{9+16}\\~\\&=&\sqrt{25}\\~\\&=&5\end{array}

\sin~\alpha=\frac{\text{AP}}{\text{OP}}~\to \red{\huge{\sin~a=\frac{4}{5}}}

\cosec~\alpha=\frac{\text{OP}}{\text{AP}}~\to \red{\huge{\cosec~a=\frac{5}{4}}}

\cos~\alpha=\frac{\text{OA}}{\text{OP}}~\to \red{\huge{\cos~a=\frac{3}{5}}}

\sec~\alpha=\frac{\text{OP}}{\text{OA}}~\to \red{\huge{\sec~a=\frac{5}{3}}}

\tan~\alpha=\frac{\text{AP}}{\text{OA}}~\to \red{\huge{\tan~a=\frac{4}{3}}}

\cot~\alpha=\frac{\text{OA}}{\text{AP}}~\to \red{\huge{\cot~a=\frac{3}{4}}}

\\

\purple{\huge{2b.}}

Perhatikan gambar keempat (terlampir) :

Dengan rumus Pythagoras :

\begin{array}{lll}\text{OP}^2&=&\text{OA}^2+\text{AP}^2\\~\\\text{OP}&=&\sqrt{\text{OA}^2+\text{AP}^2}\\~\\&=&\sqrt{1^2+\left(\sqrt{3}\right)^2}\\~\\&=&\sqrt{1+3}\\~\\&=&\sqrt{4}\\~\\&=&2\end{array}

\sin~\alpha=\frac{\text{AP}}{\text{OP}}~\to \red{\huge{\sin~a=\frac{1}{2}\sqrt{3}}}

\cosec~\alpha=\frac{\text{OP}}{\text{AP}}~\to \red{\huge{\cosec~a=\frac{2}{\sqrt{3}}=\frac{2}{3}\sqrt{3}}}

\cos~\alpha=\frac{\text{OA}}{\text{OP}}~\to \red{\huge{\cos~a=\frac{1}{2}}}

\sec~\alpha=\frac{\text{OP}}{\text{OA}}~\to \red{\huge{\sec~a=2}}

\tan~\alpha=\frac{\text{AP}}{\text{OA}}~\to \red{\huge{\tan~a=\sqrt{3}}}

\cot~\alpha=\frac{\text{OA}}{\text{AP}}~\to \red{\huge{\cot~a=\frac{1}{\sqrt{3}}=\frac{1}{3}\sqrt{3}}}

\\

\purple{\huge{3a.}}

\begin{array}{lll}\sin^2~60\degree+\cos^2~60\degree&=&\left(\frac{1}{2}\sqrt{3}\right)^2+\left(\frac{1}{2}\right)^2\\~\\&=&\frac{3}{4}+\frac{1}{4}\end{array}

\red{\huge{\sin^2~60\degree+\cos^2~60\degree=1}}

\\

\purple{\huge{3b.}}

\begin{array}{lll}\frac{2+\tan~30\degree}{1-\tan^2~30\degree}&=&\frac{2+\frac{1}{3}\sqrt{3}}{1-\left(\frac{1}{3}\sqrt{3}\right)^2}\\~\\&=&\frac{2+\frac{1}{3}\sqrt{3}}{1-\frac{1}{3}}\\~\\&=&\frac{2+\frac{1}{3}\sqrt{3}}{\frac{2}{3}}\\~\\&=&\frac{3}{2}\times \left(2+\frac{1}{3}\sqrt{3}\right)\end{array}

\red{\huge{\frac{2+\tan~30\degree}{1-\tan^2~30\degree}=3+\frac{1}{2}\sqrt{3}}}

[tex]\purple{\huge{1a.}}[/tex]Perhatikan gambar pertama (terlampir) :Dengan rumus Pythagoras :[tex]\begin{array}{lll}\text{QR}^2&=&\text{PQ}^2+\text{PR}^2\\~\\\text{PQ}^2&=&\text{QR}^2-\text{PR}^2\\~\\\text{PQ}&=&\sqrt{\text{QR}^2-\text{PR}^2}\\~\\&=&\sqrt{13^2-5^2}\\~\\&=&\sqrt{169-25}\\~\\&=&\sqrt{144}\\~\\&=&12\end{array}[/tex][tex]\sin~a=\frac{\text{PQ}}{\text{QR}}~\to \red{\huge{\sin~a=\frac{12}{13}}}[/tex][tex]\cos~a=\frac{\text{PR}}{\text{QR}}~\to \red{\huge{\cos~a=\frac{5}{13}}}[/tex][tex]\tan~a=\frac{\text{PQ}}{\text{PR}}~\to \red{\huge{\tan~a=\frac{12}{5}}}[/tex][tex]\\[/tex][tex]\purple{\huge{1b.}}[/tex]Perhatikan gambar kedua (terlampir) :Dengan rumus Pythagoras :[tex]\begin{array}{lll}\text{LM}^2&=&\text{KL}^2+\text{KM}^2\\~\\\text{LM}&=&\sqrt{\text{KL}^2+\text{KM}^2}\\~\\&=&\sqrt{7^2+2^2}\\~\\&=&\sqrt{49+4}\\~\\&=&\sqrt{53}\end{array}[/tex][tex]\sin~a=\frac{\text{KL}}{\text{LM}}~\to \red{\huge{\sin~a=\frac{7}{\sqrt{53}}=\frac{7}{53}\sqrt{53}}}[/tex][tex]\cos~a=\frac{\text{KM}}{\text{LM}}~\to \red{\huge{\cos~a=\frac{2}{\sqrt{53}}=\frac{2}{53}\sqrt{53}}}[/tex][tex]\tan~a=\frac{\text{KL}}{\text{KM}}~\to \red{\huge{\tan~a=\frac{7}{2}}}[/tex][tex]\\[/tex][tex]\purple{\huge{2a.}}[/tex]Perhatikan gambar ketiga (terlampir) :Dengan rumus Pythagoras :[tex]\begin{array}{lll}\text{OP}^2&=&\text{OA}^2+\text{AP}^2\\~\\\text{OP}&=&\sqrt{\text{OA}^2+\text{AP}^2}\\~\\&=&\sqrt{3^2+4^2}\\~\\&=&\sqrt{9+16}\\~\\&=&\sqrt{25}\\~\\&=&5\end{array}[/tex][tex]\sin~\alpha=\frac{\text{AP}}{\text{OP}}~\to \red{\huge{\sin~a=\frac{4}{5}}}[/tex][tex]\cosec~\alpha=\frac{\text{OP}}{\text{AP}}~\to \red{\huge{\cosec~a=\frac{5}{4}}}[/tex][tex]\cos~\alpha=\frac{\text{OA}}{\text{OP}}~\to \red{\huge{\cos~a=\frac{3}{5}}}[/tex][tex]\sec~\alpha=\frac{\text{OP}}{\text{OA}}~\to \red{\huge{\sec~a=\frac{5}{3}}}[/tex][tex]\tan~\alpha=\frac{\text{AP}}{\text{OA}}~\to \red{\huge{\tan~a=\frac{4}{3}}}[/tex][tex]\cot~\alpha=\frac{\text{OA}}{\text{AP}}~\to \red{\huge{\cot~a=\frac{3}{4}}}[/tex][tex]\\[/tex][tex]\purple{\huge{2b.}}[/tex]Perhatikan gambar keempat (terlampir) :Dengan rumus Pythagoras :[tex]\begin{array}{lll}\text{OP}^2&=&\text{OA}^2+\text{AP}^2\\~\\\text{OP}&=&\sqrt{\text{OA}^2+\text{AP}^2}\\~\\&=&\sqrt{1^2+\left(\sqrt{3}\right)^2}\\~\\&=&\sqrt{1+3}\\~\\&=&\sqrt{4}\\~\\&=&2\end{array}[/tex][tex]\sin~\alpha=\frac{\text{AP}}{\text{OP}}~\to \red{\huge{\sin~a=\frac{1}{2}\sqrt{3}}}[/tex][tex]\cosec~\alpha=\frac{\text{OP}}{\text{AP}}~\to \red{\huge{\cosec~a=\frac{2}{\sqrt{3}}=\frac{2}{3}\sqrt{3}}}[/tex][tex]\cos~\alpha=\frac{\text{OA}}{\text{OP}}~\to \red{\huge{\cos~a=\frac{1}{2}}}[/tex][tex]\sec~\alpha=\frac{\text{OP}}{\text{OA}}~\to \red{\huge{\sec~a=2}}[/tex][tex]\tan~\alpha=\frac{\text{AP}}{\text{OA}}~\to \red{\huge{\tan~a=\sqrt{3}}}[/tex][tex]\cot~\alpha=\frac{\text{OA}}{\text{AP}}~\to \red{\huge{\cot~a=\frac{1}{\sqrt{3}}=\frac{1}{3}\sqrt{3}}}[/tex][tex]\\[/tex][tex]\purple{\huge{3a.}}[/tex][tex]\begin{array}{lll}\sin^2~60\degree+\cos^2~60\degree&=&\left(\frac{1}{2}\sqrt{3}\right)^2+\left(\frac{1}{2}\right)^2\\~\\&=&\frac{3}{4}+\frac{1}{4}\end{array}[/tex][tex]\red{\huge{\sin^2~60\degree+\cos^2~60\degree=1}}[/tex][tex]\\[/tex][tex]\purple{\huge{3b.}}[/tex][tex]\begin{array}{lll}\frac{2+\tan~30\degree}{1-\tan^2~30\degree}&=&\frac{2+\frac{1}{3}\sqrt{3}}{1-\left(\frac{1}{3}\sqrt{3}\right)^2}\\~\\&=&\frac{2+\frac{1}{3}\sqrt{3}}{1-\frac{1}{3}}\\~\\&=&\frac{2+\frac{1}{3}\sqrt{3}}{\frac{2}{3}}\\~\\&=&\frac{3}{2}\times \left(2+\frac{1}{3}\sqrt{3}\right)\end{array}[/tex][tex]\red{\huge{\frac{2+\tan~30\degree}{1-\tan^2~30\degree}=3+\frac{1}{2}\sqrt{3}}}[/tex][tex]\purple{\huge{1a.}}[/tex]Perhatikan gambar pertama (terlampir) :Dengan rumus Pythagoras :[tex]\begin{array}{lll}\text{QR}^2&=&\text{PQ}^2+\text{PR}^2\\~\\\text{PQ}^2&=&\text{QR}^2-\text{PR}^2\\~\\\text{PQ}&=&\sqrt{\text{QR}^2-\text{PR}^2}\\~\\&=&\sqrt{13^2-5^2}\\~\\&=&\sqrt{169-25}\\~\\&=&\sqrt{144}\\~\\&=&12\end{array}[/tex][tex]\sin~a=\frac{\text{PQ}}{\text{QR}}~\to \red{\huge{\sin~a=\frac{12}{13}}}[/tex][tex]\cos~a=\frac{\text{PR}}{\text{QR}}~\to \red{\huge{\cos~a=\frac{5}{13}}}[/tex][tex]\tan~a=\frac{\text{PQ}}{\text{PR}}~\to \red{\huge{\tan~a=\frac{12}{5}}}[/tex][tex]\\[/tex][tex]\purple{\huge{1b.}}[/tex]Perhatikan gambar kedua (terlampir) :Dengan rumus Pythagoras :[tex]\begin{array}{lll}\text{LM}^2&=&\text{KL}^2+\text{KM}^2\\~\\\text{LM}&=&\sqrt{\text{KL}^2+\text{KM}^2}\\~\\&=&\sqrt{7^2+2^2}\\~\\&=&\sqrt{49+4}\\~\\&=&\sqrt{53}\end{array}[/tex][tex]\sin~a=\frac{\text{KL}}{\text{LM}}~\to \red{\huge{\sin~a=\frac{7}{\sqrt{53}}=\frac{7}{53}\sqrt{53}}}[/tex][tex]\cos~a=\frac{\text{KM}}{\text{LM}}~\to \red{\huge{\cos~a=\frac{2}{\sqrt{53}}=\frac{2}{53}\sqrt{53}}}[/tex][tex]\tan~a=\frac{\text{KL}}{\text{KM}}~\to \red{\huge{\tan~a=\frac{7}{2}}}[/tex][tex]\\[/tex][tex]\purple{\huge{2a.}}[/tex]Perhatikan gambar ketiga (terlampir) :Dengan rumus Pythagoras :[tex]\begin{array}{lll}\text{OP}^2&=&\text{OA}^2+\text{AP}^2\\~\\\text{OP}&=&\sqrt{\text{OA}^2+\text{AP}^2}\\~\\&=&\sqrt{3^2+4^2}\\~\\&=&\sqrt{9+16}\\~\\&=&\sqrt{25}\\~\\&=&5\end{array}[/tex][tex]\sin~\alpha=\frac{\text{AP}}{\text{OP}}~\to \red{\huge{\sin~a=\frac{4}{5}}}[/tex][tex]\cosec~\alpha=\frac{\text{OP}}{\text{AP}}~\to \red{\huge{\cosec~a=\frac{5}{4}}}[/tex][tex]\cos~\alpha=\frac{\text{OA}}{\text{OP}}~\to \red{\huge{\cos~a=\frac{3}{5}}}[/tex][tex]\sec~\alpha=\frac{\text{OP}}{\text{OA}}~\to \red{\huge{\sec~a=\frac{5}{3}}}[/tex][tex]\tan~\alpha=\frac{\text{AP}}{\text{OA}}~\to \red{\huge{\tan~a=\frac{4}{3}}}[/tex][tex]\cot~\alpha=\frac{\text{OA}}{\text{AP}}~\to \red{\huge{\cot~a=\frac{3}{4}}}[/tex][tex]\\[/tex][tex]\purple{\huge{2b.}}[/tex]Perhatikan gambar keempat (terlampir) :Dengan rumus Pythagoras :[tex]\begin{array}{lll}\text{OP}^2&=&\text{OA}^2+\text{AP}^2\\~\\\text{OP}&=&\sqrt{\text{OA}^2+\text{AP}^2}\\~\\&=&\sqrt{1^2+\left(\sqrt{3}\right)^2}\\~\\&=&\sqrt{1+3}\\~\\&=&\sqrt{4}\\~\\&=&2\end{array}[/tex][tex]\sin~\alpha=\frac{\text{AP}}{\text{OP}}~\to \red{\huge{\sin~a=\frac{1}{2}\sqrt{3}}}[/tex][tex]\cosec~\alpha=\frac{\text{OP}}{\text{AP}}~\to \red{\huge{\cosec~a=\frac{2}{\sqrt{3}}=\frac{2}{3}\sqrt{3}}}[/tex][tex]\cos~\alpha=\frac{\text{OA}}{\text{OP}}~\to \red{\huge{\cos~a=\frac{1}{2}}}[/tex][tex]\sec~\alpha=\frac{\text{OP}}{\text{OA}}~\to \red{\huge{\sec~a=2}}[/tex][tex]\tan~\alpha=\frac{\text{AP}}{\text{OA}}~\to \red{\huge{\tan~a=\sqrt{3}}}[/tex][tex]\cot~\alpha=\frac{\text{OA}}{\text{AP}}~\to \red{\huge{\cot~a=\frac{1}{\sqrt{3}}=\frac{1}{3}\sqrt{3}}}[/tex][tex]\\[/tex][tex]\purple{\huge{3a.}}[/tex][tex]\begin{array}{lll}\sin^2~60\degree+\cos^2~60\degree&=&\left(\frac{1}{2}\sqrt{3}\right)^2+\left(\frac{1}{2}\right)^2\\~\\&=&\frac{3}{4}+\frac{1}{4}\end{array}[/tex][tex]\red{\huge{\sin^2~60\degree+\cos^2~60\degree=1}}[/tex][tex]\\[/tex][tex]\purple{\huge{3b.}}[/tex][tex]\begin{array}{lll}\frac{2+\tan~30\degree}{1-\tan^2~30\degree}&=&\frac{2+\frac{1}{3}\sqrt{3}}{1-\left(\frac{1}{3}\sqrt{3}\right)^2}\\~\\&=&\frac{2+\frac{1}{3}\sqrt{3}}{1-\frac{1}{3}}\\~\\&=&\frac{2+\frac{1}{3}\sqrt{3}}{\frac{2}{3}}\\~\\&=&\frac{3}{2}\times \left(2+\frac{1}{3}\sqrt{3}\right)\end{array}[/tex][tex]\red{\huge{\frac{2+\tan~30\degree}{1-\tan^2~30\degree}=3+\frac{1}{2}\sqrt{3}}}[/tex][tex]\purple{\huge{1a.}}[/tex]Perhatikan gambar pertama (terlampir) :Dengan rumus Pythagoras :[tex]\begin{array}{lll}\text{QR}^2&=&\text{PQ}^2+\text{PR}^2\\~\\\text{PQ}^2&=&\text{QR}^2-\text{PR}^2\\~\\\text{PQ}&=&\sqrt{\text{QR}^2-\text{PR}^2}\\~\\&=&\sqrt{13^2-5^2}\\~\\&=&\sqrt{169-25}\\~\\&=&\sqrt{144}\\~\\&=&12\end{array}[/tex][tex]\sin~a=\frac{\text{PQ}}{\text{QR}}~\to \red{\huge{\sin~a=\frac{12}{13}}}[/tex][tex]\cos~a=\frac{\text{PR}}{\text{QR}}~\to \red{\huge{\cos~a=\frac{5}{13}}}[/tex][tex]\tan~a=\frac{\text{PQ}}{\text{PR}}~\to \red{\huge{\tan~a=\frac{12}{5}}}[/tex][tex]\\[/tex][tex]\purple{\huge{1b.}}[/tex]Perhatikan gambar kedua (terlampir) :Dengan rumus Pythagoras :[tex]\begin{array}{lll}\text{LM}^2&=&\text{KL}^2+\text{KM}^2\\~\\\text{LM}&=&\sqrt{\text{KL}^2+\text{KM}^2}\\~\\&=&\sqrt{7^2+2^2}\\~\\&=&\sqrt{49+4}\\~\\&=&\sqrt{53}\end{array}[/tex][tex]\sin~a=\frac{\text{KL}}{\text{LM}}~\to \red{\huge{\sin~a=\frac{7}{\sqrt{53}}=\frac{7}{53}\sqrt{53}}}[/tex][tex]\cos~a=\frac{\text{KM}}{\text{LM}}~\to \red{\huge{\cos~a=\frac{2}{\sqrt{53}}=\frac{2}{53}\sqrt{53}}}[/tex][tex]\tan~a=\frac{\text{KL}}{\text{KM}}~\to \red{\huge{\tan~a=\frac{7}{2}}}[/tex][tex]\\[/tex][tex]\purple{\huge{2a.}}[/tex]Perhatikan gambar ketiga (terlampir) :Dengan rumus Pythagoras :[tex]\begin{array}{lll}\text{OP}^2&=&\text{OA}^2+\text{AP}^2\\~\\\text{OP}&=&\sqrt{\text{OA}^2+\text{AP}^2}\\~\\&=&\sqrt{3^2+4^2}\\~\\&=&\sqrt{9+16}\\~\\&=&\sqrt{25}\\~\\&=&5\end{array}[/tex][tex]\sin~\alpha=\frac{\text{AP}}{\text{OP}}~\to \red{\huge{\sin~a=\frac{4}{5}}}[/tex][tex]\cosec~\alpha=\frac{\text{OP}}{\text{AP}}~\to \red{\huge{\cosec~a=\frac{5}{4}}}[/tex][tex]\cos~\alpha=\frac{\text{OA}}{\text{OP}}~\to \red{\huge{\cos~a=\frac{3}{5}}}[/tex][tex]\sec~\alpha=\frac{\text{OP}}{\text{OA}}~\to \red{\huge{\sec~a=\frac{5}{3}}}[/tex][tex]\tan~\alpha=\frac{\text{AP}}{\text{OA}}~\to \red{\huge{\tan~a=\frac{4}{3}}}[/tex][tex]\cot~\alpha=\frac{\text{OA}}{\text{AP}}~\to \red{\huge{\cot~a=\frac{3}{4}}}[/tex][tex]\\[/tex][tex]\purple{\huge{2b.}}[/tex]Perhatikan gambar keempat (terlampir) :Dengan rumus Pythagoras :[tex]\begin{array}{lll}\text{OP}^2&=&\text{OA}^2+\text{AP}^2\\~\\\text{OP}&=&\sqrt{\text{OA}^2+\text{AP}^2}\\~\\&=&\sqrt{1^2+\left(\sqrt{3}\right)^2}\\~\\&=&\sqrt{1+3}\\~\\&=&\sqrt{4}\\~\\&=&2\end{array}[/tex][tex]\sin~\alpha=\frac{\text{AP}}{\text{OP}}~\to \red{\huge{\sin~a=\frac{1}{2}\sqrt{3}}}[/tex][tex]\cosec~\alpha=\frac{\text{OP}}{\text{AP}}~\to \red{\huge{\cosec~a=\frac{2}{\sqrt{3}}=\frac{2}{3}\sqrt{3}}}[/tex][tex]\cos~\alpha=\frac{\text{OA}}{\text{OP}}~\to \red{\huge{\cos~a=\frac{1}{2}}}[/tex][tex]\sec~\alpha=\frac{\text{OP}}{\text{OA}}~\to \red{\huge{\sec~a=2}}[/tex][tex]\tan~\alpha=\frac{\text{AP}}{\text{OA}}~\to \red{\huge{\tan~a=\sqrt{3}}}[/tex][tex]\cot~\alpha=\frac{\text{OA}}{\text{AP}}~\to \red{\huge{\cot~a=\frac{1}{\sqrt{3}}=\frac{1}{3}\sqrt{3}}}[/tex][tex]\\[/tex][tex]\purple{\huge{3a.}}[/tex][tex]\begin{array}{lll}\sin^2~60\degree+\cos^2~60\degree&=&\left(\frac{1}{2}\sqrt{3}\right)^2+\left(\frac{1}{2}\right)^2\\~\\&=&\frac{3}{4}+\frac{1}{4}\end{array}[/tex][tex]\red{\huge{\sin^2~60\degree+\cos^2~60\degree=1}}[/tex][tex]\\[/tex][tex]\purple{\huge{3b.}}[/tex][tex]\begin{array}{lll}\frac{2+\tan~30\degree}{1-\tan^2~30\degree}&=&\frac{2+\frac{1}{3}\sqrt{3}}{1-\left(\frac{1}{3}\sqrt{3}\right)^2}\\~\\&=&\frac{2+\frac{1}{3}\sqrt{3}}{1-\frac{1}{3}}\\~\\&=&\frac{2+\frac{1}{3}\sqrt{3}}{\frac{2}{3}}\\~\\&=&\frac{3}{2}\times \left(2+\frac{1}{3}\sqrt{3}\right)\end{array}[/tex][tex]\red{\huge{\frac{2+\tan~30\degree}{1-\tan^2~30\degree}=3+\frac{1}{2}\sqrt{3}}}[/tex][tex]\purple{\huge{1a.}}[/tex]Perhatikan gambar pertama (terlampir) :Dengan rumus Pythagoras :[tex]\begin{array}{lll}\text{QR}^2&=&\text{PQ}^2+\text{PR}^2\\~\\\text{PQ}^2&=&\text{QR}^2-\text{PR}^2\\~\\\text{PQ}&=&\sqrt{\text{QR}^2-\text{PR}^2}\\~\\&=&\sqrt{13^2-5^2}\\~\\&=&\sqrt{169-25}\\~\\&=&\sqrt{144}\\~\\&=&12\end{array}[/tex][tex]\sin~a=\frac{\text{PQ}}{\text{QR}}~\to \red{\huge{\sin~a=\frac{12}{13}}}[/tex][tex]\cos~a=\frac{\text{PR}}{\text{QR}}~\to \red{\huge{\cos~a=\frac{5}{13}}}[/tex][tex]\tan~a=\frac{\text{PQ}}{\text{PR}}~\to \red{\huge{\tan~a=\frac{12}{5}}}[/tex][tex]\\[/tex][tex]\purple{\huge{1b.}}[/tex]Perhatikan gambar kedua (terlampir) :Dengan rumus Pythagoras :[tex]\begin{array}{lll}\text{LM}^2&=&\text{KL}^2+\text{KM}^2\\~\\\text{LM}&=&\sqrt{\text{KL}^2+\text{KM}^2}\\~\\&=&\sqrt{7^2+2^2}\\~\\&=&\sqrt{49+4}\\~\\&=&\sqrt{53}\end{array}[/tex][tex]\sin~a=\frac{\text{KL}}{\text{LM}}~\to \red{\huge{\sin~a=\frac{7}{\sqrt{53}}=\frac{7}{53}\sqrt{53}}}[/tex][tex]\cos~a=\frac{\text{KM}}{\text{LM}}~\to \red{\huge{\cos~a=\frac{2}{\sqrt{53}}=\frac{2}{53}\sqrt{53}}}[/tex][tex]\tan~a=\frac{\text{KL}}{\text{KM}}~\to \red{\huge{\tan~a=\frac{7}{2}}}[/tex][tex]\\[/tex][tex]\purple{\huge{2a.}}[/tex]Perhatikan gambar ketiga (terlampir) :Dengan rumus Pythagoras :[tex]\begin{array}{lll}\text{OP}^2&=&\text{OA}^2+\text{AP}^2\\~\\\text{OP}&=&\sqrt{\text{OA}^2+\text{AP}^2}\\~\\&=&\sqrt{3^2+4^2}\\~\\&=&\sqrt{9+16}\\~\\&=&\sqrt{25}\\~\\&=&5\end{array}[/tex][tex]\sin~\alpha=\frac{\text{AP}}{\text{OP}}~\to \red{\huge{\sin~a=\frac{4}{5}}}[/tex][tex]\cosec~\alpha=\frac{\text{OP}}{\text{AP}}~\to \red{\huge{\cosec~a=\frac{5}{4}}}[/tex][tex]\cos~\alpha=\frac{\text{OA}}{\text{OP}}~\to \red{\huge{\cos~a=\frac{3}{5}}}[/tex][tex]\sec~\alpha=\frac{\text{OP}}{\text{OA}}~\to \red{\huge{\sec~a=\frac{5}{3}}}[/tex][tex]\tan~\alpha=\frac{\text{AP}}{\text{OA}}~\to \red{\huge{\tan~a=\frac{4}{3}}}[/tex][tex]\cot~\alpha=\frac{\text{OA}}{\text{AP}}~\to \red{\huge{\cot~a=\frac{3}{4}}}[/tex][tex]\\[/tex][tex]\purple{\huge{2b.}}[/tex]Perhatikan gambar keempat (terlampir) :Dengan rumus Pythagoras :[tex]\begin{array}{lll}\text{OP}^2&=&\text{OA}^2+\text{AP}^2\\~\\\text{OP}&=&\sqrt{\text{OA}^2+\text{AP}^2}\\~\\&=&\sqrt{1^2+\left(\sqrt{3}\right)^2}\\~\\&=&\sqrt{1+3}\\~\\&=&\sqrt{4}\\~\\&=&2\end{array}[/tex][tex]\sin~\alpha=\frac{\text{AP}}{\text{OP}}~\to \red{\huge{\sin~a=\frac{1}{2}\sqrt{3}}}[/tex][tex]\cosec~\alpha=\frac{\text{OP}}{\text{AP}}~\to \red{\huge{\cosec~a=\frac{2}{\sqrt{3}}=\frac{2}{3}\sqrt{3}}}[/tex][tex]\cos~\alpha=\frac{\text{OA}}{\text{OP}}~\to \red{\huge{\cos~a=\frac{1}{2}}}[/tex][tex]\sec~\alpha=\frac{\text{OP}}{\text{OA}}~\to \red{\huge{\sec~a=2}}[/tex][tex]\tan~\alpha=\frac{\text{AP}}{\text{OA}}~\to \red{\huge{\tan~a=\sqrt{3}}}[/tex][tex]\cot~\alpha=\frac{\text{OA}}{\text{AP}}~\to \red{\huge{\cot~a=\frac{1}{\sqrt{3}}=\frac{1}{3}\sqrt{3}}}[/tex][tex]\\[/tex][tex]\purple{\huge{3a.}}[/tex][tex]\begin{array}{lll}\sin^2~60\degree+\cos^2~60\degree&=&\left(\frac{1}{2}\sqrt{3}\right)^2+\left(\frac{1}{2}\right)^2\\~\\&=&\frac{3}{4}+\frac{1}{4}\end{array}[/tex][tex]\red{\huge{\sin^2~60\degree+\cos^2~60\degree=1}}[/tex][tex]\\[/tex][tex]\purple{\huge{3b.}}[/tex][tex]\begin{array}{lll}\frac{2+\tan~30\degree}{1-\tan^2~30\degree}&=&\frac{2+\frac{1}{3}\sqrt{3}}{1-\left(\frac{1}{3}\sqrt{3}\right)^2}\\~\\&=&\frac{2+\frac{1}{3}\sqrt{3}}{1-\frac{1}{3}}\\~\\&=&\frac{2+\frac{1}{3}\sqrt{3}}{\frac{2}{3}}\\~\\&=&\frac{3}{2}\times \left(2+\frac{1}{3}\sqrt{3}\right)\end{array}[/tex][tex]\red{\huge{\frac{2+\tan~30\degree}{1-\tan^2~30\degree}=3+\frac{1}{2}\sqrt{3}}}[/tex][tex]\purple{\huge{1a.}}[/tex]Perhatikan gambar pertama (terlampir) :Dengan rumus Pythagoras :[tex]\begin{array}{lll}\text{QR}^2&=&\text{PQ}^2+\text{PR}^2\\~\\\text{PQ}^2&=&\text{QR}^2-\text{PR}^2\\~\\\text{PQ}&=&\sqrt{\text{QR}^2-\text{PR}^2}\\~\\&=&\sqrt{13^2-5^2}\\~\\&=&\sqrt{169-25}\\~\\&=&\sqrt{144}\\~\\&=&12\end{array}[/tex][tex]\sin~a=\frac{\text{PQ}}{\text{QR}}~\to \red{\huge{\sin~a=\frac{12}{13}}}[/tex][tex]\cos~a=\frac{\text{PR}}{\text{QR}}~\to \red{\huge{\cos~a=\frac{5}{13}}}[/tex][tex]\tan~a=\frac{\text{PQ}}{\text{PR}}~\to \red{\huge{\tan~a=\frac{12}{5}}}[/tex][tex]\\[/tex][tex]\purple{\huge{1b.}}[/tex]Perhatikan gambar kedua (terlampir) :Dengan rumus Pythagoras :[tex]\begin{array}{lll}\text{LM}^2&=&\text{KL}^2+\text{KM}^2\\~\\\text{LM}&=&\sqrt{\text{KL}^2+\text{KM}^2}\\~\\&=&\sqrt{7^2+2^2}\\~\\&=&\sqrt{49+4}\\~\\&=&\sqrt{53}\end{array}[/tex][tex]\sin~a=\frac{\text{KL}}{\text{LM}}~\to \red{\huge{\sin~a=\frac{7}{\sqrt{53}}=\frac{7}{53}\sqrt{53}}}[/tex][tex]\cos~a=\frac{\text{KM}}{\text{LM}}~\to \red{\huge{\cos~a=\frac{2}{\sqrt{53}}=\frac{2}{53}\sqrt{53}}}[/tex][tex]\tan~a=\frac{\text{KL}}{\text{KM}}~\to \red{\huge{\tan~a=\frac{7}{2}}}[/tex][tex]\\[/tex][tex]\purple{\huge{2a.}}[/tex]Perhatikan gambar ketiga (terlampir) :Dengan rumus Pythagoras :[tex]\begin{array}{lll}\text{OP}^2&=&\text{OA}^2+\text{AP}^2\\~\\\text{OP}&=&\sqrt{\text{OA}^2+\text{AP}^2}\\~\\&=&\sqrt{3^2+4^2}\\~\\&=&\sqrt{9+16}\\~\\&=&\sqrt{25}\\~\\&=&5\end{array}[/tex][tex]\sin~\alpha=\frac{\text{AP}}{\text{OP}}~\to \red{\huge{\sin~a=\frac{4}{5}}}[/tex][tex]\cosec~\alpha=\frac{\text{OP}}{\text{AP}}~\to \red{\huge{\cosec~a=\frac{5}{4}}}[/tex][tex]\cos~\alpha=\frac{\text{OA}}{\text{OP}}~\to \red{\huge{\cos~a=\frac{3}{5}}}[/tex][tex]\sec~\alpha=\frac{\text{OP}}{\text{OA}}~\to \red{\huge{\sec~a=\frac{5}{3}}}[/tex][tex]\tan~\alpha=\frac{\text{AP}}{\text{OA}}~\to \red{\huge{\tan~a=\frac{4}{3}}}[/tex][tex]\cot~\alpha=\frac{\text{OA}}{\text{AP}}~\to \red{\huge{\cot~a=\frac{3}{4}}}[/tex][tex]\\[/tex][tex]\purple{\huge{2b.}}[/tex]Perhatikan gambar keempat (terlampir) :Dengan rumus Pythagoras :[tex]\begin{array}{lll}\text{OP}^2&=&\text{OA}^2+\text{AP}^2\\~\\\text{OP}&=&\sqrt{\text{OA}^2+\text{AP}^2}\\~\\&=&\sqrt{1^2+\left(\sqrt{3}\right)^2}\\~\\&=&\sqrt{1+3}\\~\\&=&\sqrt{4}\\~\\&=&2\end{array}[/tex][tex]\sin~\alpha=\frac{\text{AP}}{\text{OP}}~\to \red{\huge{\sin~a=\frac{1}{2}\sqrt{3}}}[/tex][tex]\cosec~\alpha=\frac{\text{OP}}{\text{AP}}~\to \red{\huge{\cosec~a=\frac{2}{\sqrt{3}}=\frac{2}{3}\sqrt{3}}}[/tex][tex]\cos~\alpha=\frac{\text{OA}}{\text{OP}}~\to \red{\huge{\cos~a=\frac{1}{2}}}[/tex][tex]\sec~\alpha=\frac{\text{OP}}{\text{OA}}~\to \red{\huge{\sec~a=2}}[/tex][tex]\tan~\alpha=\frac{\text{AP}}{\text{OA}}~\to \red{\huge{\tan~a=\sqrt{3}}}[/tex][tex]\cot~\alpha=\frac{\text{OA}}{\text{AP}}~\to \red{\huge{\cot~a=\frac{1}{\sqrt{3}}=\frac{1}{3}\sqrt{3}}}[/tex][tex]\\[/tex][tex]\purple{\huge{3a.}}[/tex][tex]\begin{array}{lll}\sin^2~60\degree+\cos^2~60\degree&=&\left(\frac{1}{2}\sqrt{3}\right)^2+\left(\frac{1}{2}\right)^2\\~\\&=&\frac{3}{4}+\frac{1}{4}\end{array}[/tex][tex]\red{\huge{\sin^2~60\degree+\cos^2~60\degree=1}}[/tex][tex]\\[/tex][tex]\purple{\huge{3b.}}[/tex][tex]\begin{array}{lll}\frac{2+\tan~30\degree}{1-\tan^2~30\degree}&=&\frac{2+\frac{1}{3}\sqrt{3}}{1-\left(\frac{1}{3}\sqrt{3}\right)^2}\\~\\&=&\frac{2+\frac{1}{3}\sqrt{3}}{1-\frac{1}{3}}\\~\\&=&\frac{2+\frac{1}{3}\sqrt{3}}{\frac{2}{3}}\\~\\&=&\frac{3}{2}\times \left(2+\frac{1}{3}\sqrt{3}\right)\end{array}[/tex][tex]\red{\huge{\frac{2+\tan~30\degree}{1-\tan^2~30\degree}=3+\frac{1}{2}\sqrt{3}}}[/tex]

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Last Update: Thu, 26 Aug 21
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