Diketahui ∆PQR siku-siku di Q .
Besar∠P=(3x+11)° dan ∠R=(x+7)°.Besar ∠P=....
Besar∠P=(3x+11)° dan ∠R=(x+7)°.Besar ∠P=....
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3x + 11 + 90 + x + 7 = 180
4x + 108 = 180
4x = 180-108
4x = 72
x = 18
<P = 3x + 11 = 3(18) + 11 = 54 + 11 = 65
sudut R = (x=7)
sudut Q = 90 (karena siku-siku)
Besar Sudut dalam sebuat segitiga = 180
180 = sudut P + sudut Q + sudut R
180 = 3x + 11 + 90 + x + 7
180 = 4x + 108
180-108 = 4x
x = 72/4
x = 18
sudut P = (3x+11) = 3(18)+11 = 65
maaf kalo salah