Bantu kerjakan . . . . . . . .
Matematika
Dindafebria12
Pertanyaan
Bantu kerjakan
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
1 Jawaban
-
1. Jawaban whongaliem
34) BC² = AB² + AC² - 2.AC.BC cos A
7² = 9² + 8² - 2 .9 .8 cos A
49 = 81 + 64 - 144 .cos A
49 = 145 - 144.cos A
144 cos A = 145 - 49
144 cos A = 96
cos A = 96/144
= 2/3
cos² A + sin² A = 1
(2/3)² + sin² A = 1
sin² A = 1 - (2/3)²
sin² A = 1 - 4/9
sin² A = 5/9
sin A = √(5/9)
sin A = (1/3)√5 ..... jawaban : B
35) QR² = PQ² + QR² - 2.PQ .QR .cos P
= 3² + 4² - 2 . 3 . 4 cos 60°
= 9 + 16 - 24 .1/2
= 25 - 12
= 13
QR = √13
PQ² = QR² + PR² - 2.QR .PQ cos R
3² = 13 + 4² - 2 .√13 . 4 .cos R
9 = 13 + 16 - 8√13 cos R
9 = 29 - 8√13 .cos R
8√13 .cos R = 29 - 9
8√13 .cos R = 20
cos R = 20 / (8√13)
= 10/(4√13)
= (10/52)√13
= (5/26)√13 ...... jawaban : E
35) BC² = AC² + AB² - 2. AC .AB .cos A
= 6² + 2² - 2 . 6 .2 . cos 60°
= 36 + 4 - 24 . 1/2
= 40 - 12
= 28
BC = √28
AC² = BC² + AB² - 2 .BC .AB . cos B
6² = 28 + 2² - 2 .√28 . 2 .cos B
38 = 28 + 4 - 4√28 .cos B
38 = 32 - 4√28 .cos B
38 = 32 - 8√7 cos B
8√7 cos B = 32 - 38
8√7 cos B = - 6
cos B = - 6/(8 √7)
= - 3/(4√7)
= - (3√7)/28
tan² B = (1/cos² B) - 1
= {1 /(3√7 /28)²} - 1
= {1 / (63/784) } - 1
= 784 / 63 - 1
= 721 / 63
= 103/9
tan B = √(103/9)
= (1/3)√103. ....... tak terdapat dalam option