Matematika
chandrawulan02
Pertanyaan
<PQR = (4x-10)derajat
<QSR = (47 derajat)
besar < R ?
2 Jawaban
-
1. Jawaban nafinur
4x-10=47+x
4x-x=47+10
3x=57
x=19
sudut R = x
x = 19° -
2. Jawaban Milieth
∠RQS+∠PQR=180°
∠RQS=180°-(4x-10°)
∠RQS=180°-4x+10°
∠RQS=190°-4x
∠RQS+∠QSR+∠SRQ=180°
190°-4x+47°+x=180°
-3x=180° -190°- 47°
-3x=-57°
x=19°
∠R=x=19°