40.Caranya gimana? Bidang empat P.QRS dengan rusuk 5cm. jika PS=QR=2cm,maka besar sudut antara bidang PQR dan QRS adalah........ a.30⁰ P b.45⁰ c.50⁰ d.60⁰ e.75⁰
hendrisyafa
Garis tinggi PQR = PT = √PQ²-(1/2 QR)² = √5²-(1/2. 2)² = √25 - 1 = √24 = 2√6
garis tinggi QRS = ST = √QR² - (1/2 RS)² = √2² - (1/2.2)² = √4 - 1 = √3
titik berat garis ST = 1 : 2 sehingga OT : OS = 1 : 2 OT = 1/3. ST = 1/3 √3
cos α = OT / PT =
α = 83,6
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hendrisyafa
jawaban ini masih perlu dikoreksi dikarenakan soal masih perlu dikonfirmasi
= √5²-(1/2. 2)²
= √25 - 1
= √24
= 2√6
garis tinggi QRS = ST = √QR² - (1/2 RS)²
= √2² - (1/2.2)²
= √4 - 1
= √3
titik berat garis ST = 1 : 2 sehingga
OT : OS = 1 : 2
OT = 1/3. ST
= 1/3 √3
cos α = OT / PT
=
α = 83,6