Sin 150 Sin 120 Cos 210 Cos 300

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Sin 150 Sin 120 Cos 210 Cos 300
Sin 150 Sin 120 Cos 210 Cos 300

sin 150°+sin 120° per cos 210°-cos 300°​

1. sin 150°+sin 120° per cos 210°-cos 300°​

Jawab:sin(150)+sin 120/cos(210)-xos(300)

-0,7148764+sin 120/(0,88387746) + 0,022096619

-0,6927798+sin(-120/0,88387746

-0,6927798+0,6264127

-0,0663671

2. Sin 150 – sin 120 / cos 300 – cos 210

sin 150 – sin 120 / cos 300 – cos 210
= sin (180-30) – sin (180-60) / cos (360-60) – cos (180+30)
= sin 30 – sin 60 / cos 60 – (- cos 30)
= 1/2 – 1/2√3 / 1/2 + 1/2√3
= 1/2 (1 – √3) / 1/2 (1 + √3)
= (1 – √3) / (1 + √3) × (1 – √3) / (1 – √3)
= (1 – √3)(1 – √3) / (1^2 – (√3)^2)
= (1 – √3 – √3 + 3) / (1 – 3)
= (4 – 2√3) / (-2)
= 2.(2 – √3) / (-2)
= -(2 – √3)
= √3 – 2

3. Sin 150° + sin 120° Per Cos 210° – cos 300°??

kalau menurut saya ini dah jawabannya
maaf klu salah

4. Sin 150° + Sin 120° / Cos 210° -Cos 300°

Sin 150°=Sin(180-30)=Sin 30° = ½

Sin 120°=Sin(180-60)=Sin 60° = ½√3

Cos 210°=Cos(180+30)=-Cos 30°=-½√3

Cos 300°=Cos(360-60)=Cos 60°=½

Sin 150° + Sin 120° / Cos 210° -Cos 300°

=( ½+½√3) / (-½√3-½)

= ½(1+√3) / -½(√3+1)

= -1 (e)

#cmiiw

5. Nilai dari sin 150 °+ sin 120° per cos 210° + cos 300°

[tex]\frac{\sin150+\sin120}{\cos210+\cos300}\\\frac{\sin(180-30)+\sin(90+30)}{\cos(180+30)+\cos(270+30)}\\\frac{(\sin180\cos30-\cos180\sin30)+(\sin90\cos30+\cos90\sin30)}{(\cos180\cos30-\sin180\sin30)+(\cos270\cos30-\sin270\sin30)}\\\frac{(\sin30)+(\cos
30)}{(-\cos30)+(\sin30)}\\\frac{\sin30+\cos30}{-\cos30+\sin30}\\\frac{\frac{1}{2}+\frac{1}{2}\sqrt3}{-\frac{1}{2}\sqrt3+\frac{1}{2}}\\\frac{\frac{1+\sqrt3}{2}}{\frac{1-\sqrt3}{2}}\\\frac{1+\sqrt3}{2}\times\frac{2}{1-\sqrt3}\\\boxed{\frac{1+\sqrt3}{1-\sqrt3}}[/tex]

6. Sin 150° + sin 120°/cos 210° + cos 300° =…

Jawaban:

-1

(❀╹◡╹)ノ゙❀~

7. cos 210°-cos: 300⁰ sin 120° + sin 150⁰ adalah​

Jawab:

[tex]\frac{1}{4} (2-3\sqrt{3})[/tex]

Penjelasan dengan langkah-langkah:

cos 210°-cos 300⁰ sin 120° + sin 150⁰

karena semua angka kelipatan 30 maka bisa pakai tabel trigonometri, jadi;

[tex]= -\frac{1}{2}\sqrt{3} – \frac{1}{2} (\frac{1}{2}\sqrt{3} ) + \frac{1}{2} \\= -\frac{1}{2}\sqrt{3} – \frac{1}{4}\sqrt{3} + \frac{1}{2} \\=-\frac{2}{4}\sqrt{3} – \frac{1}{4}\sqrt{3} + \frac{2}{4} \\=-\frac{3}{4}\sqrt{3} + \frac{2}{4}\\=\frac{1}{4} (2-3\sqrt{3})[/tex]

8. sin 150 + sin 120 / per cos 210 – cos 300

sin(150)+sin 120/cos(210)-xos(300)
-0,7148764+sin 120/(0,88387746) + 0,022096619
-0,6927798+sin(-120/0,88387746
-0,6927798+0,6264127
-0,0663671
 aaf kalau salah

9. Nilai dari sin 150°-sin 120°/cos 300°-cos 210° adalah….

sin 150°=sin (180-30) = sin 30 = 1/2
sin 120°= sin (180 – 60)= sin 60 = √3/2
cos 300°= cos (360 -60) = cos 60=1/2
cos 210° = cos (180 + 30) =- cos 30 = -√3/2

(150°-sin 120°)/(cos 300°-cos 210°)
=(1/2 -√3/2)/(1/2+√3/2)
=(1-√3)/(1+√3)
=(1-√3)/(1+√3) ×(1-√3)/(1-√3)
=(1-2√3+3)/(1-3)
=(4-2√3)/(-2)
= √3-2

10. cos 300°+ sin 210° + tan 150° ÷ cos 210°+ sin 120° + tan 300° =

cos 300° = cos 60° = ½
sin 210° = – sin 30° = -½
tan 150° = – tan 30° = -⅓√3
cos 210° = – cos 30° = -½√3
sin 120° = sin 60° = ½√3
tan 300° = – tan 60° = -√3

(cos 300°+ sin 210° + tan 150°) / (cos 210°+ sin 120° + tan 300°)
= ( ½ – ½ – ⅓√3 ) / ( – ½√3 + ½√3 – √3 )
= ( -⅓√3 ) / ( – √3 )
= -⅓Bab Trigonometri
Matematika SMA Kelas X

(cos 300° + sin 210° + tan 150°) : (cos 210° + sin 120° + tan 300°)
= (1/2 – 1/2 – 1/3 √3) : (-1/2 √3 + 1/2 √3 + (-√3))
= (-1/3 √3) : (-√3)
= 1/3

11. Nilai dari sin 150° + sin 120°/ -cos 210° – cos 300° adalah…

Identitas

sin(180-a) = sin a

cos(180+a) = -cos a

cos(360-a) = cos a

maka

(sin 150 + sin 120)/(-cos 210 – cos 300)

= sin (180-30) + sin (180-60) / -cos(180+30) – cos (360-60)

= sin 30 + sin 60 / cos 30 – cos 60

= (1/2 + 1/2 akar(3)) / (1/2 akar(3) – 1/2)

= 1/2(1+akar(3)) / 1/2(akar(3)-1)

= 1+akar(3) / akar(3)-1

rasionalkan

= 1+akar(3) / akar(3)-1 x akar(3)+1 / akar(3)+1

= [1+akar(3)][akar(3)+1] / 3-1

= 1+2akar(3)+3 / 2

= 4+2akar(3) / 2

= 2+akar(3)

#MathIsBeautiful

12. Sin 150° + sin 120°perCOS 210“ – cos 300 °​

Jawaban:

-1

Penjelasan dengan langkah-langkah:

(sin 150° + sin 120°) / (cos 210° – cos 300°)

= (1/2 + 1/2 √3) / (-1/2 √3 – 1/2)

= (1/2 + 1/2 √3) / (-1 . (1/2 + 1/2 √3))

= 1/-1

= -1

Kode Kategorisasi : 10.2.6

Kelas 10

Pelajaran Matematika

Bab 6 – Trigonometri Dasar

13. nilai sin 150° + Sin 120°/cos 210° – cos 300°​

sin 150 + sin 120 / cos 210 – cos 330
= 1/2 + 1/2 akar 3 / -1/2 akar 3 – 1/2 akar 3
= (1 + akar 3) / 2 / – akar 3
= 1 + akar 3 / -2 akar 3
= (1 + akar 3)(2 akar 3) / -12
= 2 akar 3 + 6 / -12
= 2(akar 3 + 3) / -12
= akar 3 + 3 / -6

semoga membantu

14. Sin 150° + Sin 120° / Cos 210°-Cos 300°

Sin 150=Sin(180-30)=Sin 30° = ½

Sin 120°=Sin(180-60)=Sin 60°=½√3

Cos 210°=Cos(180+30)=-Cos 30°=-½√3

Cos 300°=Cos(360-60)=Cos 60°= ½

Sin 150° + Sin 120° / Cos 210°-Cos 300°

= ½ + ½√3 / -½√3 – ½

= ½(1+√3) / -½(√3+1)

= -1

#cmiiw

Jawaban:

sin150°sin120°=0,983633168

cos210°cos300°=0,999721562

15. nilai dari sin 150° + sin 120° / cos 210° – cos 300°​

Sin 150°=Sin(180-30)=Sin 30°=½

Sin 120°=Sin(180-60)=Sin 60°=½√3

Cos 210°=Cos(180+30)=-Cos 30°=-½√3

Cos 300°=Cos(360-60)=Cos 60°=½

Sin 150°+Sin 120°/Cos 210°-Cos 300°

= ½ + ½√3 / -½√3-½

=½(1+√3) / -½(√3+1)

= -1

#cmiiw

16. sin 150° + sin 120° cos 210° – cos 300° adalah​

sin 150° + sin 120°

= sin(180° – 30°) + sin(180° – 60°)

= sin(30°) + sin(60°)

= 1/2 + √3/2

= (1 + √3)/2

cos 120° – cos 300°

= cos(180° – 60°) – cos(360° – 60°)

= -cos(60°) – cos(60°)

= -1/2 – 1/2

= -1

17. nilai dari sin 150+sin 120 cos 210-cos 300​

~Trigonometri

sin 150° + sin 120° = sin 30° + sin 60°

sin 150° + sin 120° = ½ + ½√3

sin 150° + sin 120° = ½ (1 + √3)

cos 210° – cos 300° = – cos 30° – cos 60°

cos 210° – cos 300° = – ½√3 – ½

cos 210° – cos 300° = -½ (√3 + 1)

Maka :

(sin 150° + sin 120°) / ( cos 210° – cos 300°) = -1

18. sin 150 + sin 120cos 210 – cos 300​

~Trigono

sin 150 + sin 120 = sin 30 + sin 60

sin 150 + sin 120 = ¹/₂ + ¹/₂√3

sin 150 + sin 120 = ¹/₂ (1 + √3)

cos 210 – cos 300 = -cos 30 – cos 60

cos 210 – cos 300 = -¹/₂√3 – ¹/₂

cos 210 – cos 300 = ¹/₂ (-1 – √3)

19. Nilai dari sin 150° + sin 120° / cos 210° – cos 300°

Jawaban:

[tex]\frac{Sin150 +Sin 120}{Cos210-Cos300} =\frac{Sin30+sin60}{(-Cos30)-Cos60\\}=\frac{\frac{1}{2}+\frac{1}{2} \sqrt{3} }{\frac{-1}{2}\sqrt{3} -\frac{1}{2} } =-1[/tex]

20. sin 150° + sin 120°/cos 210° – cos 300°

1/2 + 1/2[tex] \sqrt{3[/tex] : -1/2[tex] \sqrt{3[/tex] – 1/21/2 + (1/2√3)/-1/2√3 – 1/2 = 1/2 -1 -1/2 = -1

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