ΔXYZ siku siku jika sin X = 1/5 √5 --> cos X= 2/5 (√5) sin Z = 1/10 √10 --> cos Z = 3/10 ( √10) maka akan dbuktikan apakah <(X+Z) = 90
sin (X+ Z) = sin X cos Z + cos X sin Z sin (90) = 1/5 √5 . 3/10 √10 + 2/5 √5 . 1/10 √10 1 = 3/50 √50 + 2/50 √50 1 = 5/50 √50 1= 1/10 . 5√2 1 = 1/2 √2 tidak terbukti dan < y ≠ 90
Verified answer
JawabΔXYZ siku siku
jika
sin X = 1/5 √5 --> cos X= 2/5 (√5)
sin Z = 1/10 √10 --> cos Z = 3/10 ( √10)
maka akan dbuktikan apakah <(X+Z) = 90
sin (X+ Z) = sin X cos Z + cos X sin Z
sin (90) = 1/5 √5 . 3/10 √10 + 2/5 √5 . 1/10 √10
1 = 3/50 √50 + 2/50 √50
1 = 5/50 √50
1= 1/10 . 5√2
1 = 1/2 √2
tidak terbukti
dan < y ≠ 90
Verified answer
Pembuktian ∆xyz siku"sin x = 1/√5
cos x = √(1 - sin² x) = 2/√5
sin z = 1/√10
cos z = 3/√10
sin y = sin (180 - (x + z)) = sin (x + z)
sin (x + z)
= sin x cos z + cos x sin z
= 1/√5 . 3/√10 + 2/√5 . 1/√10
= 5/√50
= 1/2 √2
1/√10 < 1/√5 < 1/√2
(1/√10)² + (1/√5)² ≠ (1/√2)²
∆xyz bukan ∆ siku"
tan y/2
= sin y / (1 + cos y)
= 1/√2 / (1 + 1/√2)
= 1/(1 + √2)
= √2 - 1