Applications in Trigonometry

Tieria

2009-08-02 4:48 am

1. The figure shows a prism whose base is a right-angled triangle. If RS = 7 cm, PR = 10 cm and UP = 9 cm, find the angle between the planes URS and PQRS.

http://i617.photobucket.com/albums/tt257/michaelcoco_/2-2.jpg

The answer is 51.6ﾟ.

2. In the figure, A, B, C and D are four vertices of a rectangular block with dimensions 5 cm x 10 cm x 12 cm. E is a point on AB such that ∠BEC = ∠AED = x. Find
(a) the measure of x,
(b) the length of DE,
(c) the angle between the lines CE and ED.

http://i617.photobucket.com/albums/tt257/michaelcoco_/3-1.jpg

The answer of (a), (b) and (c) are 51.3ﾟ, 12.8 cm and 113ﾟ respectively.

3. In the triangular pyramid as shown, TP is perpendicular to the plane PQR, and S is a point on QR such that PS ⊥ QR. If PQ = 3, PR = 4, PT = 6 and ∠QPR = 60ﾟ, find the angles between
(a) the line TS and the plane PQR.

http://i617.photobucket.com/albums/tt257/michaelcoco_/4.jpg

回答 (1)

用戶9112

2009-08-02 5:29 am

✔ 最佳答案

(1)
PR2 = RS2 + PS2
102 = 72 + PS2
PS = √(100-49) = √51
Angle between the planes URS and PQRS = X
tanX = 9 / √51 = 1.26
X = 51.6 degree
(2)(a)
tanX = CB / EB
EB = 5 / tanX
tanX = AD / AE
AE = 10 / tanX
AB = AE + EB = 12
5 / tanX + 10 / tanX = 12
tanX = 15/12 = 1.25
X = 51.3 degree
(b) sinX = AD / DE
sin51.3 = 10 / DE
DE = 10 / sin51.3 = 12.8cm
(c)sinX = CB / CE
CE = 5 / sin51.3 = 6.4m
CD2 = 52 + 102 + 122 = 269
CD = √269 = 16.4
cosCED = (CE2 + ED2 – CD2)/(2.CD.ED)
= (6.42 + 12.82 – 269)/(2*6.4*12.8)
= -0.392
CED = 113 degree
(3)
cosine rule, QR2 = QP2 + PR2 – 2.QP.PRcos60
QR2 = 9 + 16 – 2(3)(4)(1/2) = 13
QR = √13
Area of triangle = (1/2)QP.PRsin60 = (1/2)(3)(4)(√3/2) = 3√3
Area of triangle = (1/2)PS.QR = (1/2)PS(√13)
So PS = (2)(3√3)/√13) = 2.88
tanTSP = 6/2.88 = 2.08
TSP = 64.3 degree

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