關於Trigonometric Formulas的數學問題

Tieria

2009-05-31 2:42 am

1. In ΔABC, b = 5cm, c = 10.5cm and B = 25ﾟ. Find the two possible areas of ΔABC.

Ans.: 26.2cm^2, 16.0cm^2

2. In the figure, the diagonal QS divides the quadrilateral PQRS into two triangles PQS and QRS of equal areas. In ΔPQS, PS = 13cm whereas in ΔQRS, QR = 6cm, RS = 8cm and ∠QRS = 90ﾟ. Find ∠PSR.

Ans.: 58.5ﾟ

http://i617.photobucket.com/albums/tt257/michaelcoco_/3.jpg

回答 (1)

用戶10012

2009-05-31 3:21 am

✔ 最佳答案

1. By sine rule
10.5/sin C = 5/sin 25, so sin C = 2.1 sin 25 = 0.887498, so C = 62.56 or (180 - 62.56) = 117.439.
So Angle A = 180 - 25 - 62.56 = 92.44 or 180 - 25 - 117.439 = 37.561.
So Area of triangle ABC = [(10.5)(5) sin 92.44]/2 = 26.2 or
= [(10.5)(5) sin 37.561]/2 = 16.0
2.
Area of triangle QRS = (6 x8)/2 = 24 = area of triangle QPS.
Angle QSR = arctan (6/8) = 36.87..............(1)
QS = sqrt ( 6^2 + 8^2) = 10.
so area of triangle QPS = [(QS)(PS) sin (angle PSQ)]/2, that is
24 = [(10)(13) sin (angle PSQ)]/2
angle PSQ = arcsin [24/65] = 21.67................(2)
So angle PSR = (1) + (2) = 36.87 + 21.67 = 58.54 = 58.5

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