1. jika A+B = [tex] \frac{\pi}{6} [/tex] cos A cos B = 3/4 maka nilai cos (A-B)=.... a. 1/9 - 1/2 √3 b. 2/3 + 1/2 √3 c. 3/4 - 1/2 √3 d. 3/2 - 1/2 √3 e. 1/2 √3 2
Matematika
mendesdontgafos
Pertanyaan
1. jika A+B =
[tex] \frac{\pi}{6} [/tex]
cos A cos B = 3/4 maka nilai cos (A-B)=....
a. 1/9 - 1/2 √3
b. 2/3 + 1/2 √3
c. 3/4 - 1/2 √3
d. 3/2 - 1/2 √3
e. 1/2 √3
2. dik. cos (A-B) = 3/4 dan cos A cos B = 7/25
dit. tan A tan B =....
a. 8/25
b. 8/7
c. 7/8
d. -8/7
e. -8/25
tolong bantu pake cara, bsk dikumpul
[tex] \frac{\pi}{6} [/tex]
cos A cos B = 3/4 maka nilai cos (A-B)=....
a. 1/9 - 1/2 √3
b. 2/3 + 1/2 √3
c. 3/4 - 1/2 √3
d. 3/2 - 1/2 √3
e. 1/2 √3
2. dik. cos (A-B) = 3/4 dan cos A cos B = 7/25
dit. tan A tan B =....
a. 8/25
b. 8/7
c. 7/8
d. -8/7
e. -8/25
tolong bantu pake cara, bsk dikumpul
1 Jawaban
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1. Jawaban RexyGamaliel
1)
A+B = 30°
cosA.cosB = 3/4
cos(A+B) = cosA.cosB - sinA.sinB
cos 30° = 3/4 - sinA.sinB
sinA.sinB = 3/4 - 1/2 √3
cos(A-B)
= cosA.cosB + sinA.sinB
= 3/4 + 3/4 - 1/2 √3
= 3/2 - 1/2 √3
D
2)
cos(A-B) = 3/4
cosA.cosB = 7/25
cos(A-B) = cosA.cosB + sinA.sinB
3/4 = 7/25 + sinA.sinB
sinA.sinB = 47/100
tanA.tanB
= (sinA.sinB)/(cosA.cosB)
= (47/100)/(7/25)
= 47/28
*tidak ada di pilihan*