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Dear Students,
Welcome to today’s quiz! This is your opportunity to demonstrate what you’ve learned so far, so do your best. Please keep in mind that you have a maximum of 45 minutes to complete all the questions. Make sure to manage your time wisely and answer each question thoughtfully.
Good luck!
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Pertanyaan 1 dari 10
1. Pertanyaan
1 pointsJika \(\sin 25^{\circ} = \dfrac{2m}{m^2 + 1}, \: m > 1\), maka \(\sec 25^{\circ} + \cot 65^{\circ} = \dotso\)
Benar
\(\sec 65^{\circ} + \cot 65^{\circ} = \dfrac{m^2 + 1}{m^2\:-\:1} + \dfrac{2m}{m^2\:-\:1}\)
\(\sec 65^{\circ} + \cot 65^{\circ} = \dfrac{m^2 + 2m + 1}{m^2\:-\:1}\)
\(\sec 65^{\circ} + \cot 65^{\circ} = \dfrac{(m + 1)^2}{(m + 1)(m\:-\:1)}\)
\(\sec 65^{\circ} + \cot 65^{\circ} = \dfrac{m + 1}{m\:-\:1}\)
Salah
\(\sec 65^{\circ} + \cot 65^{\circ} = \dfrac{m^2 + 1}{m^2\:-\:1} + \dfrac{2m}{m^2\:-\:1}\)
\(\sec 65^{\circ} + \cot 65^{\circ} = \dfrac{m^2 + 2m + 1}{m^2\:-\:1}\)
\(\sec 65^{\circ} + \cot 65^{\circ} = \dfrac{(m + 1)^2}{(m + 1)(m\:-\:1)}\)
\(\sec 65^{\circ} + \cot 65^{\circ} = \dfrac{m + 1}{m\:-\:1}\)
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Pertanyaan 2 dari 10
2. Pertanyaan
1 pointsJika \(\cos 17^{\circ} = m\), maka nilai dari \(\tan 73^{\circ}\:-\:\sin 17^{\circ} = \dotso\)
Benar
\(\tan 73^{\circ}\:-\:\sin 17^{\circ} = \dfrac{m}{\sqrt{1\:-\:m^2}}\:-\:\dfrac{\sqrt{1\:-\:m^2}}{1}\)
\(\tan 73^{\circ}\:-\:\sin 17^{\circ} = \dfrac{m\:-\:(1\:-\:m^2)}{\sqrt{1\:-\:m^2}}\)
\(\tan 73^{\circ}\:-\:\sin 17^{\circ} = \dfrac{m^2 + m \:-\:1}{\sqrt{1\:-\:m^2}}\)
Salah
\(\tan 73^{\circ}\:-\:\sin 17^{\circ} = \dfrac{m}{\sqrt{1\:-\:m^2}}\:-\:\dfrac{\sqrt{1\:-\:m^2}}{1}\)
\(\tan 73^{\circ}\:-\:\sin 17^{\circ} = \dfrac{m\:-\:(1\:-\:m^2)}{\sqrt{1\:-\:m^2}}\)
\(\tan 73^{\circ}\:-\:\sin 17^{\circ} = \dfrac{m^2 + m \:-\:1}{\sqrt{1\:-\:m^2}}\)
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Pertanyaan 3 dari 10
3. Pertanyaan
1 pointsPada gambar di bawah ini, \(\dfrac{\sin \beta + \cos \beta}{\sec \beta\:-\:\tan \beta} = \dotso\)
Benar
\(\dfrac{\sin \beta + \cos \beta}{\sec \beta\:-\:\tan \beta} = \dfrac{\frac{3}{5} + \frac{4}{5}}{\frac{5}{4}\:-\:\frac{3}{4}}\)
\(\dfrac{\sin \beta + \cos \beta}{\sec \beta\:-\:\tan \beta} = \dfrac{\frac{7}{5}}{\frac{2}{4}}\)
\(\dfrac{\sin \beta + \cos \beta}{\sec \beta\:-\:\tan \beta} = \dfrac{28}{10}\)
\(\dfrac{\sin \beta + \cos \beta}{\sec \beta\:-\:\tan \beta} = 2,8\)
Salah
\(\dfrac{\sin \beta + \cos \beta}{\sec \beta\:-\:\tan \beta} = \dfrac{\frac{3}{5} + \frac{4}{5}}{\frac{5}{4}\:-\:\frac{3}{4}}\)
\(\dfrac{\sin \beta + \cos \beta}{\sec \beta\:-\:\tan \beta} = \dfrac{\frac{7}{5}}{\frac{2}{4}}\)
\(\dfrac{\sin \beta + \cos \beta}{\sec \beta\:-\:\tan \beta} = \dfrac{28}{10}\)
\(\dfrac{\sin \beta + \cos \beta}{\sec \beta\:-\:\tan \beta} = 2,8\)
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Pertanyaan 4 dari 10
4. Pertanyaan
1 pointsDiketahui \(\csc \alpha = 5, \: 0 < \alpha < \frac{\pi}{2}\). Maka nilai dari \(\dfrac{5\cos \alpha + 4 \sin 15^{\circ}}{\cot \alpha \:-\:4\sin 75^{\circ}}= \dotso\)
Benar
\(\color{blue} \sin 15^{\circ} = \dfrac{\sqrt{6}\:-\:\sqrt{2}}{4}\)
\(\color{blue} \sin 75^{\circ} = \dfrac{\sqrt{6} + \sqrt{2}}{4}\)
\(\dfrac{5\cos \alpha + 4 \sin 15^{\circ}}{\cot \alpha \:-\:4\sin 75^{\circ}}\)
\(\dfrac{5\left(\dfrac{2\sqrt{6}}{5}\right) + 4\left(\dfrac{\sqrt{6}}{4}\:-\:\dfrac{\sqrt{2}}{4}\right)}{\dfrac{2\sqrt{6}}{1}\:-\:4\left(\dfrac{\sqrt{6}}{4} + \dfrac{\sqrt{2}}{4}\right)}\)
\(\dfrac{2\sqrt{6} + \sqrt{6}\:-\:\sqrt{2}}{2\sqrt{6} \:-\: \sqrt{6}\:-\:\sqrt{2}}\)
\(\dfrac{3\sqrt{6}\:-\:\sqrt{2}}{\sqrt{6}\:-\:\sqrt{2}}\times \color{blue}\dfrac{\sqrt{6} + \sqrt{2}}{\sqrt{6} + \sqrt{2}}\)
\(\dfrac{16 + 4\sqrt{3}}{4}\)
\(4 + \sqrt{3}\)
Salah
\(\color{blue} \sin 15^{\circ} = \dfrac{\sqrt{6}\:-\:\sqrt{2}}{4}\)
\(\color{blue} \sin 75^{\circ} = \dfrac{\sqrt{6} + \sqrt{2}}{4}\)
\(\dfrac{5\cos \alpha + 4 \sin 15^{\circ}}{\cot \alpha \:-\:4\sin 75^{\circ}}\)
\(\dfrac{5\left(\dfrac{2\sqrt{6}}{5}\right) + 4\left(\dfrac{\sqrt{6}}{4}\:-\:\dfrac{\sqrt{2}}{4}\right)}{\dfrac{2\sqrt{6}}{1}\:-\:4\left(\dfrac{\sqrt{6}}{4} + \dfrac{\sqrt{2}}{4}\right)}\)
\(\dfrac{2\sqrt{6} + \sqrt{6}\:-\:\sqrt{2}}{2\sqrt{6} \:-\: \sqrt{6}\:-\:\sqrt{2}}\)
\(\dfrac{3\sqrt{6}\:-\:\sqrt{2}}{\sqrt{6}\:-\:\sqrt{2}}\times \color{blue}\dfrac{\sqrt{6} + \sqrt{2}}{\sqrt{6} + \sqrt{2}}\)
\(\dfrac{16 + 4\sqrt{3}}{4}\)
\(4 + \sqrt{3}\)
Hint
\(\color{blue} \sin 15^{\circ} = \dfrac{\sqrt{6}\:-\:\sqrt{2}}{4}\) \(\color{blue} \sin 75^{\circ} = \dfrac{\sqrt{6} + \sqrt{2}}{4}\) -
Pertanyaan 5 dari 10
5. Pertanyaan
1 pointsDiketahui \(\csc \alpha = 5, \:\alpha \text{ lancip}\). Nilai dari \(\dfrac{5\cos \alpha + \cot \alpha}{4\sin 75^{\circ}} = \dotso\)
Benar
\(\color{blue} \sin 75^{\circ} = \dfrac{\sqrt{6} + \sqrt{2}}{4}\)
\(\dfrac{5\cos \alpha + \cot \alpha}{4\sin 75^{\circ}}\)
\(\dfrac{5(\frac{2\sqrt{6}}{5}) + (\frac{2\sqrt{6}}{1})}{4(\frac{\sqrt{6} + \sqrt{2}}{4})}\)
\(\dfrac{4\sqrt{6}}{\sqrt{6} + \sqrt{2}} \times \color{blue} \dfrac{\sqrt{6}\:-\:\sqrt{2}}{\sqrt{6}\:-\:\sqrt{2}}\)
\(\dfrac{24\:-\:8\sqrt{3}}{4}\)
\(6\:-\:2\sqrt{3}\)
Salah
\(\color{blue} \sin 75^{\circ} = \dfrac{\sqrt{6} + \sqrt{2}}{4}\)
\(\dfrac{5\cos \alpha + \cot \alpha}{4\sin 75^{\circ}}\)
\(\dfrac{5(\frac{2\sqrt{6}}{5}) + (\frac{2\sqrt{6}}{1})}{4(\frac{\sqrt{6} + \sqrt{2}}{4})}\)
\(\dfrac{4\sqrt{6}}{\sqrt{6} + \sqrt{2}} \times \color{blue} \dfrac{\sqrt{6}\:-\:\sqrt{2}}{\sqrt{6}\:-\:\sqrt{2}}\)
\(\dfrac{24\:-\:8\sqrt{3}}{4}\)
\(6\:-\:2\sqrt{3}\)
Hint
\(\color{blue} \sin 75^{\circ} = \dfrac{\sqrt{6} + \sqrt{2}}{4}\) -
Pertanyaan 6 dari 10
6. Pertanyaan
1 pointsDiketahui \(\sec \theta = 7\), dan \(\pi < \theta < 2\pi\). Maka nilai dari \(2\tan \theta\:-\:21\sin \theta + 8\sin \frac{\pi}{3} + 10 \cos \frac{\pi}{6} = \dotso\)
Benar
Karena \(\sec \theta\) bernilai positif, maka \(\theta\) terletak di kuadran IV.
\(2\tan \theta\:-\:21\sin \theta + 8\sin \frac{\pi}{3} + 10 \cos \frac{\pi}{6}\)
\(2(-4\sqrt{3})\:-\:21(-\frac{4\sqrt{3}}{7}) + 8(\frac{\sqrt{3}}{2}) + 10(\frac{\sqrt{3}}{2})\)
\(-8\sqrt{3} + 12\sqrt{3} + 4\sqrt{3} + 5\sqrt{3}\)
\(13\sqrt{3}\)
Salah
Karena \(\sec \theta\) bernilai positif, maka \(\theta\) terletak di kuadran IV.
\(2\tan \theta\:-\:21\sin \theta + 8\sin \frac{\pi}{3} + 10 \cos \frac{\pi}{6}\)
\(2(-4\sqrt{3})\:-\:21(-\frac{4\sqrt{3}}{7}) + 8(\frac{\sqrt{3}}{2}) + 10(\frac{\sqrt{3}}{2})\)
\(-8\sqrt{3} + 12\sqrt{3} + 4\sqrt{3} + 5\sqrt{3}\)
\(13\sqrt{3}\)
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Pertanyaan 7 dari 10
7. Pertanyaan
1 pointsNilai dari \(\dfrac{\tan 45^{\circ} + \tan 60^{\circ} }{2\cos 30^{\circ}\:-\:\cos 0^{\circ} \:-\:\sin 90^{\circ}} = \dotso\)
Benar
\(\dfrac{\tan 45^{\circ} + \tan 60^{\circ} }{2\cos 30^{\circ}\:-\:\cos 0^{\circ} \:-\:\sin 90^{\circ}}\)
\(\dfrac{1 + \sqrt{3}}{2(\frac{1}{2}\sqrt{3})\:-\:1\:-\:1}\)
\(\dfrac{1 + \sqrt{3}}{\sqrt{3}\:-\:2} \times \color{blue} \dfrac{\sqrt{3} + 2}{\sqrt{3} + 2}\)
\(\dfrac{3\sqrt{3} + 5}{-1}\)
\(-3\sqrt{3}\:-\:5\)
Salah
\(\dfrac{\tan 45^{\circ} + \tan 60^{\circ} }{2\cos 30^{\circ}\:-\:\cos 0^{\circ} \:-\:\sin 90^{\circ}}\)
\(\dfrac{1 + \sqrt{3}}{2(\frac{1}{2}\sqrt{3})\:-\:1\:-\:1}\)
\(\dfrac{1 + \sqrt{3}}{\sqrt{3}\:-\:2} \times \color{blue} \dfrac{\sqrt{3} + 2}{\sqrt{3} + 2}\)
\(\dfrac{3\sqrt{3} + 5}{-1}\)
\(-3\sqrt{3}\:-\:5\)
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Pertanyaan 8 dari 10
8. Pertanyaan
1 pointsNilai dari \(\dfrac{6\cdot \cos \frac{\pi}{3}\cdot \cos \frac{\pi}{4}\:-\:6\cdot\sin \frac{\pi}{3}\cdot \sin \frac{\pi}{4}}{\cos \frac{\pi}{3} + \tan \frac{\pi}{4}}= \dotso\)
Benar
\(\dfrac{6\cdot \cos \frac{\pi}{3}\cdot \cos \frac{\pi}{4}\:-\:6\cdot \sin \frac{\pi}{3}\cdot \sin \frac{\pi}{4}}{\cos \frac{\pi}{3} + \tan \frac{\pi}{4}}\)
\(\dfrac{6\cdot \frac{1}{2} \cdot \frac{1}{2} \sqrt{2}\:-\:6\cdot \frac{1}{2}\sqrt{3}\cdot \frac{1}{2}\sqrt{2}}{\frac{1}{2} + 1}\)
\(\dfrac{\frac{6}{4}\sqrt{2}\:-\:\frac{6}{4}\sqrt{6}}{\frac{3}{2}}\)
\(\frac{6}{4}\cdot \frac{2}{3}\cdot (\sqrt{2}\:-\:\sqrt{6})\)
\(\sqrt{2}\:-\:\sqrt{6}\)
Salah
\(\dfrac{6\cdot \cos \frac{\pi}{3}\cdot \cos \frac{\pi}{4}\:-\:6\cdot \sin \frac{\pi}{3}\cdot \sin \frac{\pi}{4}}{\cos \frac{\pi}{3} + \tan \frac{\pi}{4}}\)
\(\dfrac{6\cdot \frac{1}{2} \cdot \frac{1}{2} \sqrt{2}\:-\:6\cdot \frac{1}{2}\sqrt{3}\cdot \frac{1}{2}\sqrt{2}}{\frac{1}{2} + 1}\)
\(\dfrac{\frac{6}{4}\sqrt{2}\:-\:\frac{6}{4}\sqrt{6}}{\frac{3}{2}}\)
\(\frac{6}{4}\cdot \frac{2}{3}\cdot (\sqrt{2}\:-\:\sqrt{6})\)
\(\sqrt{2}\:-\:\sqrt{6}\)
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Pertanyaan 9 dari 10
9. Pertanyaan
1 pointsTerdapat beberapa pernyataan:
(1) \(\sin 150^{\circ} > \sin 190^{\circ}\)
(2) \(\cos 170^{\circ} < \tan 222^{\circ}\)
(3) \(\sin 100^{\circ} < \sin 200^{\circ}\)
(4) \(\sec 290^{\circ} > \sec 150^{\circ}\)
Banyaknya pernyataan yang benar ada sebanyak …
Benar
(1) \(\sin 150^{\circ}\color{red} (+) \color{black} > \sin 190^{\circ} \color{red} (-) \text{ benar}\)
(2) \(\cos 170^{\circ} \color{red} (-) \color{black} < \tan 222^{\circ}\color{red} (+) \text{ benar}\)
(3) \(\sin 100^{\circ} \color{red} (+) \color{black} < \sin 200^{\circ}\color{red} (-) \text{ salah}\)
(4) \(\sec 290^{\circ} \color{red} (+) \color{black} > \sec 150^{\circ}\color{red} (-) \text{ benar}\)
Banyaknya pernyataan yang benar ada sebanyak 3
Salah
(1) \(\sin 150^{\circ}\color{red} (+) \color{black} > \sin 190^{\circ} \color{red} (-) \text{ benar}\)
(2) \(\cos 170^{\circ} \color{red} (-) \color{black} < \tan 222^{\circ}\color{red} (+) \text{ benar}\)
(3) \(\sin 100^{\circ} \color{red} (+) \color{black} < \sin 200^{\circ}\color{red} (-) \text{ salah}\)
(4) \(\sec 290^{\circ} \color{red} (+) \color{black} > \sec 150^{\circ}\color{red} (-) \text{ benar}\)
Banyaknya pernyataan yang benar ada sebanyak 3
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Pertanyaan 10 dari 10
10. Pertanyaan
1 pointsDiketahui \(\tan \text{ M} = -1,875, \: \sin \text{ N} = 0,6\), M dan N adalah sudut tumpul. Nilai dari \(\dfrac{\sin \text{ M}\:-\:\cos \text{ N}}{\cos \text{ M} + \sin \text{ N}} = \dotso\)
Benar
\(\tan \text{ M} = -1.875 = -\dfrac{15}{8}\)
\(\sin \text{ N} = 0,6 = \dfrac{6}{10}\)
\(\dfrac{\sin \text{ M}\:-\:\cos \text{ N}}{\cos \text{ M} + \sin \text{ N}}\)
\(\dfrac{\frac{15}{17}\:-\:(-\frac{8}{10})}{(-\frac{8}{17}) + \frac{6}{10}}\)
\(\dfrac{\frac{143}{85}}{\frac{11}{85}}\)
\(\frac{143}{\cancel{85}}\times \frac{\cancel{85}}{11}\)
\(13\)
Salah
\(\tan \text{ M} = -1.875 = -\dfrac{15}{8}\)
\(\sin \text{ N} = 0,6 = \dfrac{6}{10}\)
\(\dfrac{\sin \text{ M}\:-\:\cos \text{ N}}{\cos \text{ M} + \sin \text{ N}}\)
\(\dfrac{\frac{15}{17}\:-\:(-\frac{8}{10})}{(-\frac{8}{17}) + \frac{6}{10}}\)
\(\dfrac{\frac{143}{85}}{\frac{11}{85}}\)
\(\frac{143}{\cancel{85}}\times \frac{\cancel{85}}{11}\)
\(13\)