Tolong ya yg bisa bantu saya kerjakan sama cara nya. trimakasih
DB45
Cos θ = 2/3 = y/x 2x = 3y y = ²/₃ x pitagoras --> x² = y²+5² x² = (²/₃ x)² + 25 x² =⁴/₉x² + 25 ....kalikan 9 9x² = 4x² + 225 5x² = 225 x²= 45 x = √45 x = 3√5
13. ΔKLM siku siku di L KL = √20 = 2√5 sin M = 2/3 --> KL/KM = 2/3 KM = 3/2 KL KM = (3/2) (2√5) KM = 3√5
14. ΔDFE siku siku di F EF = 3
Luas = 9 1/2 (DF x EF) = 9 DF x EF = 18 (3) DF = 18 DF = 6
DE² = EF²+ DF² DE² = 9+36 = 45 DE = 3√5
Cos E = EF/DE cos E = 3/(3√5) cos E = ¹/₅ √5
15. sin α = AD/DB --> DB = AD/sin α .....(1) cos β = BC/DB --> DB = BC/cos β.....(2) . (1) = (2) AD/sin α = BC/cos β BC = (AD. cos β)/ sin α BC = (p. cos β)/ (sin α) ...
2x = 3y
y = ²/₃ x
pitagoras --> x² = y²+5²
x² = (²/₃ x)² + 25
x² =⁴/₉x² + 25 ....kalikan 9
9x² = 4x² + 225
5x² = 225
x²= 45
x = √45
x = 3√5
13. ΔKLM siku siku di L
KL = √20 = 2√5
sin M = 2/3 --> KL/KM = 2/3
KM = 3/2 KL
KM = (3/2) (2√5)
KM = 3√5
14. ΔDFE siku siku di F
EF = 3
Luas = 9
1/2 (DF x EF) = 9
DF x EF = 18
(3) DF = 18
DF = 6
DE² = EF²+ DF²
DE² = 9+36 = 45
DE = 3√5
Cos E = EF/DE
cos E = 3/(3√5)
cos E = ¹/₅ √5
15.
sin α = AD/DB --> DB = AD/sin α .....(1)
cos β = BC/DB --> DB = BC/cos β.....(2)
.
(1) = (2)
AD/sin α = BC/cos β
BC = (AD. cos β)/ sin α
BC = (p. cos β)/ (sin α)
...