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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 30 minutes to complete all the questions. Make sure to manage your time wisely and answer each question thoughtfully.
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Question 1 of 20
1. Question
1 pointsPerhatikan gambar di bawah ini:
Jika panjang 1 kotak adalah 1 satuan, maka \(\sin \beta = \dotso\)Correct
\(\sin \beta = \dfrac{\text{depan}}{\text{miring}}\)
\(\sin \beta = \dfrac{3}{\sqrt{13}}\)
\(\sin \beta = \dfrac{3}{13}\sqrt{13}\)
Incorrect
\(\sin \beta = \dfrac{\text{depan}}{\text{miring}}\)
\(\sin \beta = \dfrac{3}{\sqrt{13}}\)
\(\sin \beta = \dfrac{3}{13}\sqrt{13}\)
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Question 2 of 20
2. Question
1 pointsJika titik A\((m, n)\), O\((0, 0)\), OA = \(r\) dan OA dengan sumbu \(x\) positif membentuk sudut \(\alpha\), maka \(\cos \alpha = \dotso\)
Correct
\(\cos \alpha = \dfrac{\text{samping}}{\text{miring}}\)
\(\cos \alpha = \dfrac{m}{r}\)
Incorrect
\(\cos \alpha = \dfrac{\text{samping}}{\text{miring}}\)
\(\cos \alpha = \dfrac{m}{r}\)
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Question 3 of 20
3. Question
1 pointsUntuk sudut α dan β pada gambar di bawah ini, maka pernyataan berikut yang benar adalah …
Correct
Semua fungsi trigonometri di kuadran I bernilai positif.
Tangen di kuadran III bernilai positif.
Incorrect
Semua fungsi trigonometri di kuadran I bernilai positif.
Tangen di kuadran III bernilai positif.
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Question 4 of 20
4. Question
1 points\(1\dfrac{3}{4}\pi \text{ rad} + 3\dfrac{1}{5}\pi \text{ rad} = \dotso ^{\circ}\)
Correct
\(1\dfrac{3}{4}\pi \text{ rad} + 3\dfrac{1}{5}\pi \text{ rad}\)
\(\dfrac{7}{4}\pi\times \dfrac{180^{\circ}}{\pi} + \dfrac{16}{5}\pi\times \dfrac{180^{\circ}}{\pi}\)
\(315^{\circ} + 576^{\circ}\)
\(891^{\circ}\)
Incorrect
\(1\dfrac{3}{4}\pi \text{ rad} + 3\dfrac{1}{5}\pi \text{ rad}\)
\(\dfrac{7}{4}\pi\times \dfrac{180^{\circ}}{\pi} + \dfrac{16}{5}\pi\times \dfrac{180^{\circ}}{\pi}\)
\(315^{\circ} + 576^{\circ}\)
\(891^{\circ}\)
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Question 5 of 20
5. Question
1 pointsSebuah eskalator memiliki panjang 34 meter dan tinggi 17 meter menghubungkan lantai dasar dengan lantai 1 di sebuah stasiun kereta api. Jika β adalah sudut yang dibentuk antara eskalator dengan lantai dasar, maka nilai β adalah …
Correct
\(\sin \beta = \dfrac{\text{depan}}{\text{miring}}\)
\(\sin \beta = \dfrac{17}{34} = \dfrac{1}{2}\)
\(\beta = 30^{\circ}\)
Incorrect
\(\sin \beta = \dfrac{\text{depan}}{\text{miring}}\)
\(\sin \beta = \dfrac{17}{34} = \dfrac{1}{2}\)
\(\beta = 30^{\circ}\)
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Question 6 of 20
6. Question
1 pointsJika \(\csc \beta = \dfrac{37}{35}\) maka nilai \(\tan \beta = \dotso\)
Correct
\(\csc \beta = \dfrac{37}{35} = \dfrac{\text{miring}}{\text{depan}}\)
\(\tan \beta = \dfrac{\text{depan}}{\text{samping}} = \dfrac{35}{12}\)
Incorrect
\(\csc \beta = \dfrac{37}{35} = \dfrac{\text{miring}}{\text{depan}}\)
\(\tan \beta = \dfrac{\text{depan}}{\text{samping}} = \dfrac{35}{12}\)
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Question 7 of 20
7. Question
1 pointsSeseorang melakukan perjalanan dengan rute A – B – C – D – E seperti pada gambar di bawah ini:
Jika \(y\) merupakan perbedaan ketinggian tempat antara A dan E, maka nilai \(y\) adalah …
(\(\sin 23^{\circ} = 0,4)\)
Correct
Perhatikan gambar berikut:
\(y = m + n\)
\(y = 2\sin 23^{\circ} + 5\sin 23^{\circ}\)
\(y = 2(0,4) + 5(0, 4)\)
\(y = 0,8 + 2\)
\(y = 2,8 \text{ km}\)Jadi perbedaan ketinggian antara tempat A dan E adalah 2,8 km
Incorrect
Perhatikan gambar berikut:
\(y = m + n\)
\(y = 2\sin 23^{\circ} + 5\sin 23^{\circ}\)
\(y = 2(0,4) + 5(0, 4)\)
\(y = 0,8 + 2\)
\(y = 2,8 \text{ km}\)Jadi perbedaan ketinggian antara tempat A dan E adalah 2,8 km
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Question 8 of 20
8. Question
1 points\(\dfrac{\cos 30^{\circ}\times \csc 60^{\circ}}{\tan 30^{\circ}} = \dotso\)
Correct
\(\dfrac{\frac{1}{2}\sqrt{3}\times \frac{2}{\sqrt{3}}}{\frac{1}{\sqrt{3}}} = \sqrt{3}\)
Incorrect
\(\dfrac{\frac{1}{2}\sqrt{3}\times \frac{2}{\sqrt{3}}}{\frac{1}{\sqrt{3}}} = \sqrt{3}\)
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Question 9 of 20
9. Question
1 pointsNilai dari \(\sin 45^{\circ}\cdot \cos 60^{\circ} + \cos 45^{\circ}\cdot \sin 60^{\circ} = \dotso\)
Correct
\(\sin 45^{\circ}\cdot \cos 60^{\circ} + \cos 45^{\circ}\cdot \sin 60^{\circ}\)
\(\frac{1}{2}\sqrt{2}\cdot \frac{1}{2} +\frac{1}{2}\sqrt{2}\cdot \frac{1}{2}\sqrt{3}\)
\(\frac{1}{4}\sqrt{2} +\frac{1}{4}\sqrt{6}\)
\(\dfrac{\sqrt{2} + \sqrt{6}}{4}\)
Incorrect
\(\sin 45^{\circ}\cdot \cos 60^{\circ} + \cos 45^{\circ}\cdot \sin 60^{\circ}\)
\(\frac{1}{2}\sqrt{2}\cdot \frac{1}{2} +\frac{1}{2}\sqrt{2}\cdot \frac{1}{2}\sqrt{3}\)
\(\frac{1}{4}\sqrt{2} +\frac{1}{4}\sqrt{6}\)
\(\dfrac{\sqrt{2} + \sqrt{6}}{4}\)
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Question 10 of 20
10. Question
1 pointsNilai dari \(\sec \frac{1}{4}\pi \cdot \cot \frac{1}{3}\pi + \sqrt{6}\cdot \csc \frac{1}{2}\pi = \dotso\)
Correct
\(\sec \frac{1}{4}\pi \cdot \cot \frac{1}{3}\pi + \sqrt{6}\cdot \csc \frac{1}{2}\pi\)
\(\sqrt{2}\cdot \frac{1}{3}\sqrt{3} + \sqrt{6}\cdot 1\)
\(\frac{1}{3}\sqrt{6} + \sqrt{6}\)
\(\frac{4}{3}\sqrt{6}\)
Incorrect
\(\sec \frac{1}{4}\pi \cdot \cot \frac{1}{3}\pi + \sqrt{6}\cdot \csc \frac{1}{2}\pi\)
\(\sqrt{2}\cdot \frac{1}{3}\sqrt{3} + \sqrt{6}\cdot 1\)
\(\frac{1}{3}\sqrt{6} + \sqrt{6}\)
\(\frac{4}{3}\sqrt{6}\)
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Question 11 of 20
11. Question
1 pointsJika titik P\((x, 14)\) adalah sebuah titik di kuadran II, O adalah titik (0, 0), dan β adalah sudut yang dibentuk oleh sumbu \(x\) positif dengan garis OP, serta \(\sin \beta = \dfrac{7}{25}\), maka nilai \(x\) adalah …
Correct
Panjang OP (sisi miring) = \(\sqrt{x^2 + 14^2}\)
\(\sin \beta = \dfrac{\text{depan}}{\text{miring}}\)
\(\dfrac{7}{25} = \dfrac{14}{\sqrt{x^2 + 196}}\)
\(\dfrac{\cancel{7}}{25} = \dfrac{\cancelto{2}{14}}{\sqrt{x^2 + 196}}\)
\(50 = \sqrt{x^2 + 196}\)
\(2500 = x^2 + 196\)
\(x^2 = 2500\:-\:196\)
\(x^2 = 2304\)
\(x = \pm\sqrt{2304}\)
\(x = \pm 48\)
Karena titik P berada di kuadran II, maka nilai \(x\) haruslah bertanda negatif, jadi \(x = -48\)
Incorrect
Panjang OP (sisi miring) = \(\sqrt{x^2 + 14^2}\)
\(\sin \beta = \dfrac{\text{depan}}{\text{miring}}\)
\(\dfrac{7}{25} = \dfrac{14}{\sqrt{x^2 + 196}}\)
\(\dfrac{\cancel{7}}{25} = \dfrac{\cancelto{2}{14}}{\sqrt{x^2 + 196}}\)
\(50 = \sqrt{x^2 + 196}\)
\(2500 = x^2 + 196\)
\(x^2 = 2500\:-\:196\)
\(x^2 = 2304\)
\(x = \pm\sqrt{2304}\)
\(x = \pm 48\)
Karena titik P berada di kuadran II, maka nilai \(x\) haruslah bertanda negatif, jadi \(x = -48\)
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Question 12 of 20
12. Question
1 pointsJika titik Q\((x, 5)\) adalah sebuah titik yang terletak di kuadran I, O adalah titik (0, 0), dan α adalah sudut yang dibentuk oleh garis OQ dengan sumbu \(x\) positif, serta \(\sin \alpha = \dfrac{15}{17}\), maka nilai \(x\) adalah …
Correct
Panjang OQ (sisi miring) = \(\sqrt{x^2 + 5^2}\)
\(\sin \alpha = \dfrac{\text{depan}}{\text{miring}}\)
\(\dfrac{\cancelto{3}{15}}{17} = \dfrac{\cancel{5}}{\sqrt{x^2 + 25}}\)
\(3\sqrt{x^2 + 25} = 17\)
Kuadratkan kedua ruas,
\(9(x^2 + 25) = 289\)
\(9x^2 + 225 = 289\)
\(9x^2 = 289\:-\:225\)
\(9x^2 = 64\)
\(x^2 = \dfrac{64}{9}\)
\(x = \pm \sqrt{\dfrac{64}{9}}\)
\(x = \pm \dfrac{8}{3}\)
Karena \(x\) terletak di kuadran I, maka nilai \(x = \dfrac{8}{3}\)
Incorrect
Panjang OQ (sisi miring) = \(\sqrt{x^2 + 5^2}\)
\(\sin \alpha = \dfrac{\text{depan}}{\text{miring}}\)
\(\dfrac{\cancelto{3}{15}}{17} = \dfrac{\cancel{5}}{\sqrt{x^2 + 25}}\)
\(3\sqrt{x^2 + 25} = 17\)
Kuadratkan kedua ruas,
\(9(x^2 + 25) = 289\)
\(9x^2 + 225 = 289\)
\(9x^2 = 289\:-\:225\)
\(9x^2 = 64\)
\(x^2 = \dfrac{64}{9}\)
\(x = \pm \sqrt{\dfrac{64}{9}}\)
\(x = \pm \dfrac{8}{3}\)
Karena \(x\) terletak di kuadran I, maka nilai \(x = \dfrac{8}{3}\)
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Question 13 of 20
13. Question
1 pointsPerhatikan gambar di bawah ini:
Nilai \(\csc \alpha = \dotso\)
Correct
Panjang OP (sisi miring) = \(\sqrt{4^2 + 2^2}\)
Panjang OP (sisi miring) = \(\sqrt{20} = 2\sqrt{5}\)
\(\csc \alpha = \dfrac{\text{miring}}{\text{depan}}\)
\(\csc \alpha = -\dfrac{2\sqrt{5}}{2}\)
\(\csc \alpha = -\sqrt{5}\)
\(\csc \alpha\) di kuadran III bernilai negatif
Incorrect
Panjang OP (sisi miring) = \(\sqrt{4^2 + 2^2}\)
Panjang OP (sisi miring) = \(\sqrt{20} = 2\sqrt{5}\)
\(\csc \alpha = \dfrac{\text{miring}}{\text{depan}}\)
\(\csc \alpha = -\dfrac{2\sqrt{5}}{2}\)
\(\csc \alpha = -\sqrt{5}\)
\(\csc \alpha\) di kuadran III bernilai negatif
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Question 14 of 20
14. Question
1 pointsPerhatikan gambar di bawah ini:
Nilai \(cot \beta = \dotso\)
Correct
\(\cot \beta = \dfrac{\text{samping}}{\text{depan}}\)
\(\cot \beta = -\dfrac{3}{3} = -1\)
Incorrect
\(\cot \beta = \dfrac{\text{samping}}{\text{depan}}\)
\(\cot \beta = -\dfrac{3}{3} = -1\)
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Question 15 of 20
15. Question
1 pointsDalam segitiga siku-siku ABC, besar sudut B = \(90^{\circ}\), \(\cos \text{A} = \dfrac{12}{13}\), dan panjang sisi BC = 25 cm. Luas segitiga ABC adalah …
Correct
Perhatikan gambar:
Luas segitiga ABC = ½ × AB × BC
Luas segitiga ABC = ½ × 60 cm × 25 cm
Luas segitiga ABC = 750 cm²
Incorrect
Perhatikan gambar:
Luas segitiga ABC = ½ × AB × BC
Luas segitiga ABC = ½ × 60 cm × 25 cm
Luas segitiga ABC = 750 cm²
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Question 16 of 20
16. Question
1 pointsPerhatikan gambar berikut:
Jika panjang sisi DC = \(m\), maka panjang sisi AD = …
Correct
Lihat segitiga siku-siku DBC
\(\sin \alpha = \dfrac{\text{DB}}{\text{DC}}\)
\(\sin \alpha = \dfrac{\text{DB}}{m}\)
\(\text{DB} = m\cdot \sin \alpha\)
Lihat segitiga siku-siku BAD
\(\cos \beta = \dfrac{\text{AD}}{\text{DB}}\)
\(\cos \beta = \dfrac{\text{AD}}{m\cdot \sin \alpha}\)
\(\text{AD} = m\cdot \sin \alpha\cdot \cos \beta\)
Incorrect
Lihat segitiga siku-siku DBC
\(\sin \alpha = \dfrac{\text{DB}}{\text{DC}}\)
\(\sin \alpha = \dfrac{\text{DB}}{m}\)
\(\text{DB} = m\cdot \sin \alpha\)
Lihat segitiga siku-siku BAD
\(\cos \beta = \dfrac{\text{AD}}{\text{DB}}\)
\(\cos \beta = \dfrac{\text{AD}}{m\cdot \sin \alpha}\)
\(\text{AD} = m\cdot \sin \alpha\cdot \cos \beta\)
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Question 17 of 20
17. Question
1 pointsKoordinat kartesius titik \((2\sqrt{3}, 60^{\circ})\) adalah …
Correct
\(x = r\cdot \cos \theta\)
\(x = 2\sqrt{3}\cdot \frac{1}{2}\)
\(x = \sqrt{3}\)
\(y = r\cdot \sin \theta\)
\(y = 2\sqrt{3}\cdot \frac{1}{2}\sqrt{3}\)
\(y = 3\)
Jadi koordinat kartesiusnya adalah \((\sqrt{3}, 3)\)
Incorrect
\(x = r\cdot \cos \theta\)
\(x = 2\sqrt{3}\cdot \frac{1}{2}\)
\(x = \sqrt{3}\)
\(y = r\cdot \sin \theta\)
\(y = 2\sqrt{3}\cdot \frac{1}{2}\sqrt{3}\)
\(y = 3\)
Jadi koordinat kartesiusnya adalah \((\sqrt{3}, 3)\)
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Question 18 of 20
18. Question
1 pointsKoordinat polar dari titik \((9, 3\sqrt{3})\) adalah …
Correct
\(r = \sqrt{x^2 + y^2}\)
\(r = \sqrt{9^2 + (3\sqrt{3})^2}\)
\(r = \sqrt{81 + 27}\)
\(r = \sqrt{108}\)
\(r = 6\sqrt{3}\)
\(\tan \theta = \dfrac{y}{x}\)
\(\tan \theta = \dfrac{3\sqrt{3}}{9}\)
\(\tan \theta = \dfrac{1}{3}\sqrt{3}\)
\(\theta = 30^{\circ}\)
Jadi koordinat polarnya adalah \((6\sqrt{3}, 30^{\circ})\)
Incorrect
\(r = \sqrt{x^2 + y^2}\)
\(r = \sqrt{9^2 + (3\sqrt{3})^2}\)
\(r = \sqrt{81 + 27}\)
\(r = \sqrt{108}\)
\(r = 6\sqrt{3}\)
\(\tan \theta = \dfrac{y}{x}\)
\(\tan \theta = \dfrac{3\sqrt{3}}{9}\)
\(\tan \theta = \dfrac{1}{3}\sqrt{3}\)
\(\theta = 30^{\circ}\)
Jadi koordinat polarnya adalah \((6\sqrt{3}, 30^{\circ})\)
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Question 19 of 20
19. Question
1 pointsJika koordinat kutub titik \((m, 3)\) adalah \((r, 45^{\circ})\), maka nilai \(m + r = \dotso\)
Correct
\(y = r\cdot \sin \theta\)
\(3 = r \cdot \sin 45^{\circ}\)
\(3 = r \cdot \frac{1}{2}\sqrt{2}\)
\(r = \dfrac{3}{\frac{\sqrt{2}}{2}}\)
\(r = \dfrac{6}{\sqrt{2}}\times \color{red} \dfrac{\sqrt{2}}{\sqrt{2}}\)
\(r = 3\sqrt{2}\)
\(x = r\cdot \cos \theta\)
\(m = 3\sqrt{2} \cdot \cos 45^{\circ}\)
\(m = 3\sqrt{2} \cdot \frac{1}{2}\sqrt{2}\)
\(m = 3\)
Jadi nilai \(m + r = 3 + 3\sqrt{2}\)
Incorrect
\(y = r\cdot \sin \theta\)
\(3 = r \cdot \sin 45^{\circ}\)
\(3 = r \cdot \frac{1}{2}\sqrt{2}\)
\(r = \dfrac{3}{\frac{\sqrt{2}}{2}}\)
\(r = \dfrac{6}{\sqrt{2}}\times \color{red} \dfrac{\sqrt{2}}{\sqrt{2}}\)
\(r = 3\sqrt{2}\)
\(x = r\cdot \cos \theta\)
\(m = 3\sqrt{2} \cdot \cos 45^{\circ}\)
\(m = 3\sqrt{2} \cdot \frac{1}{2}\sqrt{2}\)
\(m = 3\)
Jadi nilai \(m + r = 3 + 3\sqrt{2}\)
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Question 20 of 20
20. Question
1 pointsPerhatikan gambar berikut!
Jika \(\cos \alpha = \frac{2}{3}\), panjang AC = 21 cm, dan besar sudut BCD = 45°, maka panjang BC adalah …
Correct
Lihat segitiga siku-siku ADC,
\(\cos \alpha = \dfrac{\text{AD}}{\text{AC}}\)
\(\dfrac{2}{3} = \dfrac{\text{AD}}{21 \text{ cm}}\)
\(\text{AD} = \dfrac{42}{3} = 14 \text{ cm}\)
\(\text{DC} = \sqrt{(21)^2 \:-\: (14)^2}\)
\(\text{DC} = \sqrt{441 \:-\: 196}\)
\(\text{DC} = \sqrt{245}\)
\(\text{DC} = 7\sqrt{5}\text{ cm}\)
Lihat segitiga siku-siku BDC,
\(\sin \angle \text{BCD} = \dfrac{\text{DC}}{\text{BC}}\)
\(\sin 45^{\circ} = \dfrac{7\sqrt{5}}{\text{BC}}\)
\(\dfrac{\sqrt{2}}{2} = \dfrac{7\sqrt{5}}{\text{BC}}\)
\(2\cdot 7\sqrt{5} = \sqrt{2}\text{BC}\)
\(\dfrac{14\sqrt{5}}{\sqrt{2}} = \text{BC}\)
\(\text{BC} = \dfrac{14\sqrt{5}}{\sqrt{2}}\times \color{red} \dfrac{\sqrt{2}}{\sqrt{2}}\)
\(\text{BC} = \dfrac{14\sqrt{10}}{2} = 7\sqrt{10}\text{ cm}\)
Incorrect
Lihat segitiga siku-siku ADC,
\(\cos \alpha = \dfrac{\text{AD}}{\text{AC}}\)
\(\dfrac{2}{3} = \dfrac{\text{AD}}{21 \text{ cm}}\)
\(\text{AD} = \dfrac{42}{3} = 14 \text{ cm}\)
\(\text{DC} = \sqrt{(21)^2 \:-\: (14)^2}\)
\(\text{DC} = \sqrt{441 \:-\: 196}\)
\(\text{DC} = \sqrt{245}\)
\(\text{DC} = 7\sqrt{5}\text{ cm}\)
Lihat segitiga siku-siku BDC,
\(\sin \angle \text{BCD} = \dfrac{\text{DC}}{\text{BC}}\)
\(\sin 45^{\circ} = \dfrac{7\sqrt{5}}{\text{BC}}\)
\(\dfrac{\sqrt{2}}{2} = \dfrac{7\sqrt{5}}{\text{BC}}\)
\(2\cdot 7\sqrt{5} = \sqrt{2}\text{BC}\)
\(\dfrac{14\sqrt{5}}{\sqrt{2}} = \text{BC}\)
\(\text{BC} = \dfrac{14\sqrt{5}}{\sqrt{2}}\times \color{red} \dfrac{\sqrt{2}}{\sqrt{2}}\)
\(\text{BC} = \dfrac{14\sqrt{10}}{2} = 7\sqrt{10}\text{ cm}\)