sin 1 2 π \frac{1}{2}\pi 2 1 π . sin 1 3 π \frac{1}{3}\pi 3 1 π . sin 1 6 π \frac{1}{6}\pi 6 1 π =...

1 4 3 \frac{1}{4}\sqrt{3} 4 1 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

1 2 3 \frac{1}{2}\sqrt{3} 2 1 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

1 2 2 \frac{1}{2}\sqrt{2} 2 1 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

1 4 2 \frac{1}{4}\sqrt{2} 4 1 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

sin 60 ° 60\degree 6 0 ° cos 30 ° 30\degree 3 0 ° - cos 60 ° 60\degree 6 0 ° sin 30 ° 30\degree 3 0 ° =...

1 2 3 \frac{1}{2}\sqrt{3} 2 1 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

1 2 2 \frac{1}{2}\sqrt{2} 2 1 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

Sebuah tangga yang panjangnya 2 m disandarkan ke bagian paling atas sebuah tembok. Jika tinggi tembok = 3 \sqrt{3} 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> m dan α \alpha α merupakan sudut yang dibentuk oleh tangga dan tembok, maka α \alpha α =...

5 12 π \frac{5}{12}\pi 1 2 5 π

Jika sin A = 1 2 3 \ \frac{1}{2}\sqrt{3} 2 1 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> dan A lancip, maka tan A = ...

3 \sqrt{3} 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

1 3 3 \frac{1}{3}\sqrt{3} 3 1 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

1 2 3 \frac{1}{2}\sqrt{3} 2 1 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

Untuk 0 ° ≤ x ≤ 360 ° 0\degree\le x\le360\degree 0 ° ≤ x ≤ 3 6 0 ° dalam persamaan 2 sin ⁡ x − 2 = 0 2\sin x-\sqrt{2}=0 2 sin x − 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> = 0 mempunyai himpunan penyelesaian ...

{ 45 ° , 135 ° } \left\{45\degree,135\degree\right\} { 4 5 ° , 1 3 5 ° }

{ 45 ° , 315 ° } \left\{45\degree,315\degree\right\} { 4 5 ° , 3 1 5 ° }

{ 135 ° , 225 ° } \left\{135\degree,225\degree\right\} { 1 3 5 ° , 2 2 5 ° }

{ 45 ° , 135 ° , 225 ° } \left\{45\degree,135\degree,225\degree\right\} { 4 5 ° , 1 3 5 ° , 2 2 5 ° }

{ 45 ° , 135 ° , 225 ° , 315 ° } \left\{45\degree,135\degree,225\degree,315\degree\right\} { 4 5 ° , 1 3 5 ° , 2 2 5 ° , 3 1 5 ° }

Penyelesaian persamaan sin ⁡ x = 1 2 2 \sin\ x=\frac{1}{2}\sqrt{2\ } sin x = 2 1 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> untuk 0 ° ≤ x ≤ 360 ° 0\degree\le x\le360\degree 0 ° ≤ x ≤ 3 6 0 ° adalah . . .

x = 45 ° , 135 ° x=45\degree,\ 135\degree x = 4 5 ° , 1 3 5 °

x = 45 ° , 225 ° x=45\degree,\ 225\degree x = 4 5 ° , 2 2 5 °

x = 45 ° , 315 ° x=45\degree,\ 315\degree x = 4 5 ° , 3 1 5 °

x = 135 ° , 225 ° x=135\degree,\ 225\degree x = 1 3 5 ° , 2 2 5 °

x = 225 ° , 315 ° x=225\degree,\ 315\degree x = 2 2 5 ° , 3 1 5 °

diketahui cos 2x = cos 6 0 o 60^o 6 0 o , tentukan nilai x yang memenuhi untuk 0 o ≤ x ≤ 18 0 o 0^o\le x\le180^o 0 o ≤ x ≤ 1 8 0 o

3 0 o d a n 15 0 o 30^odan\ 150^o 3 0 o d a n 1 5 0 o

1 5 o d a n 3 0 o 15^odan\ 30^o 1 5 o d a n 3 0 o

6 0 o d a n 12 0 o 60^odan\ 120^o 6 0 o d a n 1 2 0 o

9 0 o d a n 18 0 o 90^o\ dan\ 180^o 9 0 o d a n 1 8 0 o

18 0 o d a n 36 0 o 180^odan\ 360^o 1 8 0 o d a n 3 6 0 o

Himpunan penyelesaian persamaan sinx = sin 3 0 ° 30^{\degree} 3 0 ° pada interval 0 ° ≤ x ≤ 36 0 ° 0^{\degree}\le x\le360^{\degree} 0 ° ≤ x ≤ 3 6 0 ° adalah...

30 ° 30\degree 3 0 ° dan 15 0 ° 150^{\degree} 1 5 0 °

3 0 ° 30^{\degree} 3 0 ° dan 60 ° 60\degree 6 0 °

30 ° 30\degree 3 0 ° dan 90 ° 90\degree 9 0 °

30 ° 30\degree 3 0 ° dan 12 0 ° 120^{\degree} 1 2 0 °

3 0 ° 30^{\degree} 3 0 ° dan 21 0 ° 210^{\degree} 2 1 0 °

Himpunan penyelesaian dari tan ⁡ ⁡ ( x + 25 ° ) = tan ⁡ ⁡ 50 ° \tan⁡(x+25°)=\tan⁡50° tan ⁡ ( x + 2 5 ° ) = tan ⁡ 5 0 ° untuk 0 ° ≤ x ≤ 360 ° 0°≤x≤360° 0 ° ≤ x ≤ 3 6 0 ° adalah . . .

{ 25 ° , 205 ° } \left\{25°,205°\right\} { 2 5 ° , 2 0 5 ° }

{ 25 ° , 105 ° } \left\{25°,105°\right\} { 2 5 ° , 1 0 5 ° }

{ 25 ° , 305 ° } \left\{25°,305°\right\} { 2 5 ° , 3 0 5 ° }

{ 75 ° , 105 ° } \left\{75°,105°\right\} { 7 5 ° , 1 0 5 ° }

{ 75 ° , 205 ° } \left\{75°,205°\right\} { 7 5 ° , 2 0 5 ° }

2 cos ⁡ ( x − π 3 ) = 3 2\cos\left(x-\frac{\pi}{3}\right)=\sqrt{3} 2 cos ( x − 3 π ) = 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> untuk 0 ° ≤ x ≤ 2 π 0\degree\le x\le2\pi 0 ° ≤ x ≤ 2 π mempunyai Himpunan penyelesaian ....

{ π 6 , π 2 } \left\{\frac{\pi}{6},\frac{\pi}{2}\right\} { 6 π , 2 π }

{ π 6 , π 3 } \left\{\frac{\pi}{6},\frac{\pi}{3}\right\} { 6 π , 3 π }

{ π 3 , π 2 } \left\{\frac{\pi}{3},\frac{\pi}{2}\right\} { 3 π , 2 π }

{ π 6 , 3 2 π } \left\{\frac{\pi}{6},\frac{3}{2}\pi\right\} { 6 π , 2 3 π }

{ π 3 , 3 2 π } \left\{\frac{\pi}{3},\frac{3}{2}\pi\right\} { 3 π , 2 3 π }

Trigonometri

Authored by Nesy Widyastuti

Mathematics

2nd Grade

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MULTIPLE SELECT QUESTION

30 sec • 1 pt

Manakah pernyataan berikut yang salah ....

\cos\alpha=\frac{samping}{miring}

$\tan\alpha=\frac{depan}{samping}$

$\operatorname{cotan}\alpha=\frac{samping}{miring}$

$\sin\alpha=\frac{depan}{miring}$

$\sec\alpha=\frac{miring}{depan}$

MULTIPLE SELECT QUESTION

30 sec • 1 pt

Nilai x pada interval $0\degree\le x\le360\degree$ yang memenuhi persamaan tan x = - 1

$0\degree$

$360\degree$

$315\degree$

$270\degree$

$135\degree$

MULTIPLE CHOICE QUESTION

30 sec • 5 pts

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Keliling suatu segienam beraturan adalah 72 cm. Luas segi enam tersebut adalah ...

432 √3 cm²

432 cm²

216 √3 cm²

216 √2 cm²

216 cm²

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

$\tan\propto=...$

$\frac{3}{4}$

$\frac{4}{3}$

$\frac{4}{5}$

$\frac{3}{5}$

$\frac{5}{4}$

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Pernyataan yang benar berikut ini adalah ....

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

sin $\frac{1}{2}\pi$ . sin $\frac{1}{3}\pi$ . sin $\frac{1}{6}\pi$ =...

$\frac{1}{2}\sqrt{3}$

$\frac{1}{2}\sqrt{2}$

$\frac{1}{4}\sqrt{3}$

$\frac{1}{4}\sqrt{2}$

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Trigonometri

Manakah pernyataan berikut yang salah ....

Nilai x pada interval $0\degree\le x\le360\degree$ yang memenuhi persamaan tan x = - 1

Keliling suatu segienam beraturan adalah 72 cm. Luas segi enam tersebut adalah ...

$\tan\propto=...$

Pernyataan yang benar berikut ini adalah ....

sin 150^o di kuadran

sin $\frac{1}{2}\pi$ . sin $\frac{1}{3}\pi$ . sin $\frac{1}{6}\pi$ =...

Nilai dari Sin 240^o + Sin 225^o + Cos 315^o = ….

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Similar Resources on Wayground

Popular Resources on Wayground

Discover more resources for Mathematics

Trigonometri

Manakah pernyataan berikut yang salah ....

Nilai x pada interval 0°≤x≤360°0\degree\le x\le360\degree0°≤x≤360° yang memenuhi persamaan tan x = - 1

Keliling suatu segienam beraturan adalah 72 cm. Luas segi enam tersebut adalah ...

tan⁡∝=...\tan\propto=...tan∝=...

Pernyataan yang benar berikut ini adalah ....

sin 150o di kuadran

sin 12π\frac{1}{2}\pi21​π . sin 13π\frac{1}{3}\pi31​π . sin 16π\frac{1}{6}\pi61​π =...

Nilai dari Sin 240o + Sin 225o + Cos 315o = ….

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Popular Resources on Wayground

Discover more resources for Mathematics

Nilai x pada interval $0\degree\le x\le360\degree$ yang memenuhi persamaan tan x = - 1

$\tan\propto=...$

sin 150^o di kuadran

sin $\frac{1}{2}\pi$ . sin $\frac{1}{3}\pi$ . sin $\frac{1}{6}\pi$ =...

Nilai dari Sin 240^o + Sin 225^o + Cos 315^o = ….