Tolong bantu please A-G pake cara kolo gk bisa jawaban nya saya butuh caranya !
Matematika
Kyuhyun45
Pertanyaan
Tolong bantu please A-G pake cara kolo gk bisa jawaban nya saya butuh caranya !
1 Jawaban
-
1. Jawaban Ricoam216
a.
x = cos 45 x 12
= [tex] \frac{1}{ \sqrt{2} } [/tex] x 12
=[tex] \frac{12}{ \sqrt{2} } [/tex] x [tex] \frac{\sqrt{2}}{ \sqrt{2} } [/tex]
= [tex] \frac{12\sqrt{2}}{ 2 } [/tex]
= [tex] 6\sqrt{2}[/tex]
b.
x = sin 30 x 16
= 0,5 x 16
= 8
c.
9 = sin 60 . X
= [tex] \frac{ \sqrt{3} }{2} [/tex] . X
X = 9 : [tex] \frac{ \sqrt{3} }{2} [/tex]
= 9 x [tex] \frac{2 }{\sqrt{3} } [/tex]
= [tex] \frac{18 }{\sqrt{3} } [/tex] x [tex] \frac{\sqrt{3} }{\sqrt{3} } [/tex]
= [tex] \frac{18\sqrt{3} }{3} [/tex]
= [tex] 6\sqrt{3} [/tex]
d.
40 = tan 30 . X
= [tex] \frac{1}{ \sqrt{3} } [/tex] . X
X = 40 : [tex] \frac{1}{ \sqrt{3} } [/tex]
= 40 x [tex]\sqrt{3}[/tex]
= [tex] 40\sqrt{3} [/tex]
e.
6 = tan 30 . a
= [tex] \frac{1}{ \sqrt{3} } [/tex] . a
a = 6 : [tex] \frac{1}{ \sqrt{3} } [/tex]
= 6 x [tex] \sqrt{3}[/tex]
= [tex] 6 \sqrt{3}[/tex]
6 = tan 60 . b
= [tex] \sqrt{3}[/tex] . b
b = 6 : [tex] \sqrt{3}[/tex]
= [tex]\frac{6}{ \sqrt{3} } [/tex] [tex] \frac{\sqrt{3}}{ \sqrt{3} }[/tex]
= [tex] \frac{6\sqrt{3}}{ 3 }[/tex]
= [tex] 2\sqrt{3}[/tex]
f.
misalkan y adalah garis yang di tengah
5 = cos 60 . y
= 1/2 . y
y = 5 : 1/2
= 5 x 2
= 10
10 = sin 60 . X
= [tex] \frac{ \sqrt{3} }{2} [/tex] . X
X = 10 : [tex] \frac{ \sqrt{3} }{2} [/tex]
= 10 x [tex] \frac{ 2 }{\sqrt{3} } [/tex]
= [tex] \frac{ 20 }{\sqrt{3} } [/tex] x [tex] \frac{ \sqrt{3}}{\sqrt{3} } [/tex]
= [tex] \frac{ 20\sqrt{3}}{3 } [/tex]
g.
misalkan sisi miring = X + Y
sisi miring = C
y = cos 30 . 24
= [tex] \frac{ \sqrt{3} }{2} [/tex] . 24
= [tex] 12 \sqrt{3} [/tex]
24 = cos 30 . C
= [tex] \frac{ \sqrt{3} }{2} [/tex] . C
C = 24 : [tex] \frac{ \sqrt{3} }{2} [/tex]
= 24 x [tex] \frac{2}{ \sqrt{3}} [/tex]
= [tex] \frac{48}{ \sqrt{3}} [/tex] x [tex] \frac{ \sqrt{3}}{ \sqrt{3}} [/tex]
= [tex] \frac{ 48\sqrt{3}}{ 3} [/tex]
= [tex] 16\sqrt{3} [/tex]
C = X + Y
[tex] 16\sqrt{3} [/tex] = X + [tex] 12\sqrt{3} [/tex]
X = [tex] 16\sqrt{3} [/tex] - [tex] 12\sqrt{3} [/tex]
= [tex] 4\sqrt{3} [/tex]
semoga membantu:)