急!F5 Solving Trinages 10q3

26.
(a)
cosB + cosD
= 2cos[(B + D)/2]cos[(B - D)/2]
= 2cos90°cos[(B - D)/2]
= 2(0)cos[(B - D)/2]
= 0

(b)

cosB
= (AB^2 + BC^2 - AC^2)/2(AB)(BC)
= (16 + 16 - AC^2)/2(4)(4)
= 1 - (AC^2)/32

cosD
= (AD^2 + DC^2 - AC^2)/2(AD)(DC)
= (9 + 36 - AC^2)/2(3)(6)
= 5/4 - (AC^2)/36

i.e.
1 - (AC^2)/32 + 5/4 - (AC^2)/36 = 0
(AC^2)(1/32 + 1/36) = 9/4
(AC^2)(17/288) = 9/4
AC^2 = 2592/68 = 648/17 = (18^2)(2/17)
AC = 18(2/17)^(1/2) cm , since AC > 0

37.
(a)
A = (1/2)20PQ ...(i)
A = (1/2)12QR ...(ii)
A = (1/2)15RP ...(iii)
By (i), (ii), (iii):
PQ = A/10, QR = A/6, RP = 2A/15

(b)
s
= (1/2)(PQ + QR + RP)
= (1/2)(A/10 + A/6 + 2A/15)
= (1/2)(3A + 5A + 4A)/30
= A/5

(c)
A
= (1/2)(PQ)(PR)sin(P)
where
sin(P)
= [1 - cos^2(P)]^(1/2)
and
cos(P)
= (PQ^2 + PR^2 - QR^2)/2(PQ)(PR)

hence
A
= (1/2)(PQ)(PR){1 - [(PQ^2 + PR^2 - QR^2)/2(PQ)(PR)]^2}^(1/2)
= (1/4){[2(PQ)(PR)]^2 - (PQ^2 + PR^2 - QR^2)^2}^(1/2)
= (1/4){[2(A/10)(A/6)]^2 - [(A^2)/100 + (A^2)/36 - 4(A^2)/225]^2}^(1/2) , by (a)
= (1/4){[(A^2)/30]^2 - [(A^2)/50]^2}^(1/2)
= (1/4)[(A^4)(1/900 - 1/2500)]^(1/2)
= (1/4)[(A^4)(16/22500)]^(1/2)
= (1/4)(4/150)A^2
= (A^2)/150
A^2 - 150A = 0
A(A - 150) = 0
A = 150 or A = 0 (rejected)
A = 150
i.e. area of triangle PQR = 150 cm^2

(d)
by (a),
PQ = 150/10 = 15 cm
QR = 150/6 = 25 cm
RP = 2(150)/15 = 20 cm

2012-12-15 05:47:26 補充：
For part(c) in question 37,
You may directly apply Heron's formula if it is allowed:
A
= [s(s - PQ)(s - QR)(s - RP)]^(1/2)
= [(A/5)(A/5 - A/10)(A/5 - A/6)(A/5 - 2A/15)]^(1/2)
= [(A/5)(A/10)(A/30)(A/15)]^(1/2)
= (A^2)[1/5(10)(30)(15)]^(1/2)
= (A^2)(1/22500)^(1/2)
= (A^2)/150