C o s 270 ° = ⋯ Cos\ 270°=⋯ C o s 2 7 0 ° = ⋯

− 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 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">

T a n 300 ° = ⋯ Tan\ 300°=⋯ T a n 3 0 0 ° = ⋯

− 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">

Nilai dari sin ⁡ 45 ° + cos ⁡ 225 ° \sin\ 45\degree+\cos\ 225\degree sin 4 5 ° + cos 2 2 5 ° adalah...

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 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">

Diketahui tan ⁡ A = − 12 5 \tan A=-\frac{12}{5} tan A = − 5 1 2 dan 270 ° ≤ A ≤ 360 ° 270\degree\le A\le360\degree 2 7 0 ° ≤ A ≤ 3 6 0 ° . Nilai cos A = ...

− 13 5 -\frac{13}{5} − 5 1 3

− 5 13 -\frac{5}{13} − 1 3 5

Sin ( − 210 ° ) \left(-210\degree\right) ( − 2 1 0 ° ) = ...

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 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">

sin ⁡ 540 ° + sin ⁡ 180 ° = . . . \sin\ 540\degree+\sin\ 180\degree=... sin 5 4 0 ° + sin 1 8 0 ° = . . .

− 2 -\sqrt{2} − 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 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">

2 \sqrt{2} 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">

4 sin ⁡ 60 ° . cos ⁡ ( − 300 ° ) − tan ⁡ 420 ° = . . . . 4\ \sin60\degree.\cos\left(-300\degree\right)-\tan\ 420\degree=.... 4 sin 6 0 ° . cos ( − 3 0 0 ° ) − tan 4 2 0 ° = . . . .

− 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 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">

N i l a i sin ⁡ 30 ° tan ⁡ 135 ° sin ⁡ 20 ° cos ⁡ 60 ° sin ⁡ 225 ° cos ⁡ 110 ° = . . . Nilai\frac{\sin30\degree\tan\ 135\degree\sin20\degree}{\cos60\degree\sin225\degree\cos110\degree}=... N i l a i cos 6 0 ° sin 2 2 5 ° cos 1 1 0 ° sin 3 0 ° tan 1 3 5 ° sin 2 0 ° = . . .

− 2 -\sqrt{2} − 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">

2 \sqrt{2} 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">

cos ⁡ ( 270 − A ) − sin ⁡ ( 180 − A ) + tan ⁡ ( 270 − A ) = . . . \cos\left(270-A\right)-\sin\left(180-A\right)+\tan\left(270-A\right)=... cos ( 2 7 0 − A ) − sin ( 1 8 0 − A ) + tan ( 2 7 0 − A ) = . . .

− 2 sin ⁡ A + cotan ⁡ A -2\sin A+\operatorname{cotan}A − 2 sin A + c o t a n A

− 2 sin ⁡ A − tan ⁡ A -2\sin A-\tan A − 2 sin A − tan A

− 2 cos ⁡ A + tan ⁡ A -2\cos A+\tan A − 2 cos A + tan A

− 2 cos ⁡ A − cotan ⁡ A -2\cos A-\operatorname{cotan}A − 2 cos A − c o t a n A

− 2 sin ⁡ A − cotan ⁡ A -2\sin A-\operatorname{cotan}A − 2 sin A − c o t a n A

Sudut berelasi

Authored by Lutviarini Latifah

Mathematics

10th Grade

CCSS covered

Used 39+ times

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25 questions

Show all answers

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

$Cos\ 270°=⋯$

$-\frac{1}{2}$

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

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

- 1

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

$Tan\ 300°=⋯$

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

- 1

$-\sqrt{3}$

$\sim$

Tags

CCSS.HSF.TF.A.2

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Nilai dari $\sin\ 45\degree+\cos\ 225\degree$ adalah...

$0$

$1$

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

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

$-1$

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

Nilai tan (-30)° = ...

⅓√3

√3

-1

-√3

-⅓√3

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Diketahui sin α bernilai negatif, cos α bernilai positif, maka α adalah sudut di kuadran …..

III

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Diketahui $\tan A=-\frac{12}{5}$ dan $270\degree\le A\le360\degree$ . Nilai cos A = ...

$-\frac{13}{5}$

$-\frac{5}{13}$

$\frac{5}{12}$

$\frac{5}{13}$

$\frac{13}{5}$

MULTIPLE CHOICE QUESTION

1 min • 1 pt

$Nilai\ dari\ \cos\ 210^0\ +\ \sin\ 300^0\ =\ .\ .\ .$

$-\sqrt{3}$

$-1$

$0$

$\frac{1}{2}$

$1$

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

$\frac{\sin270\degree.\cos135\degree.\tan135\degree}{\sin150\degree.\cos225\degree}=...$

$-2$

$1$

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

$2$

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

Tags

CCSS.HSF.TF.C.9

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Sin $\left(-210\degree\right)$ = ...

$\frac{1}{2}$

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

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

$-\frac{1}{2}$

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

Tags

CCSS.HSF.TF.A.4

10.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

$\sin\ 540\degree+\sin\ 180\degree=...$

$-1$

$-\sqrt{2}$

$0$

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

$\sqrt{2}$

Tags

CCSS.HSF.TF.C.9

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Sudut berelasi

Cos 270°=⋯Cos\ 270°=⋯Cos 270°=⋯

Tan 300°=⋯Tan\ 300°=⋯Tan 300°=⋯

Nilai dari sin⁡ 45°+cos⁡ 225°\sin\ 45\degree+\cos\ 225\degreesin 45°+cos 225° adalah...

Nilai tan (-30)° = ...

Diketahui sin α bernilai negatif, cos α bernilai positif, maka α adalah sudut di kuadran …..

Diketahui tan⁡A=−125\tan A=-\frac{12}{5}tanA=−512​ dan 270°≤A≤360°270\degree\le A\le360\degree270°≤A≤360° . Nilai cos A = ...

Nilai dari cos⁡ 2100 + sin⁡ 3000 = . . .Nilai\ dari\ \cos\ 210^0\ +\ \sin\ 300^0\ =\ .\ .\ .Nilai dari cos 2100 + sin 3000 = . . .

sin⁡270°.cos⁡135°.tan⁡135°sin⁡150°.cos⁡225°=...\frac{\sin270\degree.\cos135\degree.\tan135\degree}{\sin150\degree.\cos225\degree}=...sin150°.cos225°sin270°.cos135°.tan135°​=...

Sin (−210°)\left(-210\degree\right)(−210°) = ...

sin⁡ 540°+sin⁡ 180°=...\sin\ 540\degree+\sin\ 180\degree=...sin 540°+sin 180°=...

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$Cos\ 270°=⋯$

$Tan\ 300°=⋯$

Nilai dari $\sin\ 45\degree+\cos\ 225\degree$ adalah...

Diketahui $\tan A=-\frac{12}{5}$ dan $270\degree\le A\le360\degree$ . Nilai cos A = ...

$Nilai\ dari\ \cos\ 210^0\ +\ \sin\ 300^0\ =\ .\ .\ .$

$\frac{\sin270\degree.\cos135\degree.\tan135\degree}{\sin150\degree.\cos225\degree}=...$

Sin $\left(-210\degree\right)$ = ...

$\sin\ 540\degree+\sin\ 180\degree=...$