tolong dijawab dngan caranya bsk mw kumpl

27) luas ∆ = 1/2 . PQ . QR . sin Q
= 1/2 . 15 . 20 . sin 150°
= 1/2 . 15 . 20 . 1/2
= 15 . 5
= 75 cm^2

28) luas ∆ = 1/2 . PR . QR . sin R
= 1/2 . 8 . 9 sin 60°
= 1/2 . 8 . 9 . 1/2 √3
= 18√3 cm^2

29) tinggi tiang = 12 m => sisi depan
Misal jarak A ke tiang = x dan jarak B ke tiang = y
Maka jarak A ke B = y - x
Tan A = 12/x
Tan 45° = 12/x
1 = 12/x
x = 12

Tan B = 12/y
Tan 30° = 12/y
1/√3 = 12/y
y = 12√3

Jarak A ke B = y - x = 12√3 - 12 = 12(√3 - 1) m

27)

L = 1/2 × PQ × QR × sin 150°
= 1/2 × 15 × 20 × 1/2
= 75 cm^2 (A)

28)

L = 1/2 × PR × QR × sin 60°
= 1/2 × 8 × 9 × 1/2 akar 3
= (18 akar 3) cm^2 (E)

29)

tinggi tiang bendera (T) = 12 m
< A = 45°
< B = 30°
misal :
jarak tiang ke sudut A (45°) = a
jarak tiang ke sudut B (30°) = b
Tan A = T / a
Tan 45° = 12 / a
a = 12 / tan 45°
a = 12 / 1
a = 12 m
Tan B = T / b
Tan 30° = 12 / b
b = 12 / tan 30°
b = 12 / 1/3 akar 3
b = 12× 3 / akar 3
b = (36/akar 3)(akar 3/akar 3)
b = (36 akar 3) / 3
b = (12 akar 3) m
jarak A ke B = b - a
= 12 akar 3 - 12
= 12(akar 3 - 1) m (E)