數學不懂做-三角學

Q.16(a)
I guess you mean "Express r in terms of q" (not p).
∠R + 55º + 28º = 180º (∠ sum of △)
∠R = 97º
Using sine formula,
r / sin 97º = q / sin 55º
r = q sin 97º / sin 55º

Q.16(b)
Area of △ = ½ q r sin∠P = 20
½ (r sin 55º / sin 97º) r sin 28º = 20
r² = 40 sin 97º / (sin 55º sin 28º) ≈ 103.2374
r ≈ 10.1606 = 10.2 cm (to 3 sig. fig.)

Q.17(a)
Area of △PQR
= ½ (15) (21) sin 68º
≈ 146.0315 = 146 cm² (to 3 sig. fig.)

Q.17(b)
I guess you mean "S is a point on PR such that QS bisects ∠PQR".
Area of △PQR
　= ½ (21) (QS) sin 34º + ½ (15) (QS) sin 68º = 146.0315
18 (QS) sin 34º = 146.0315
QS = 146.0315 / (18 sin 34º)
　 ≈ 14.5082 = 14.5 cm (to 3 sig. fig.)

--- answer finished ---

2007-02-05 21:23:11 補充：
I suggest you post your 2 questions in two separate threads.It is too time-consuming to answer two questions in a single thread for just 5 points.

16a)
∠R=180-55-28=97 (△內角和)

根據正弦公式，

r / sin97 = q / sin 55

r = q sin 97 / sin 55

b)
0.5 x r x q x sin28 = 20

r x ( r / sin97 ) x sin 55 = 40 / sin 28

r ^ 2 = 103.237360

r = 10.1605= 10.2 cm

17a)

0.5 x 15 x 21 x sin 68 = 146.031 = 146 cm^2

b)
不懂