F.5-MATH(Triangles)

1.
50° + 60° + C = 180° (∠ sum of Δ)
C = 70°

The area of ΔABC :
(1/2)•a•b•sin70° = 82 cm² ...... [1]
(1/2)•b•c•sin50° = 82 cm² ...... [2]
(1/2)•c•a•sin60° = 82 cm² ...... [3]

[1] * [2] * [3] :
(1/8)•a²•b²•c²•sin70°•sin50°•sin60°= 82³ cm⁶ ...... [4]

[4]/[1]² :
(1/2)•c²•sin50°•sin60°/sin70°= 82 cm²
c = 15.24 cm

[4]/[2]² :
(1/2)•a²•sin70°•sin60°/sin50°= 82 cm²
a = 12.42 cm

[4]/[2]² :
(1/2)•b²•sin70°•sin50°/sin60°= 82 cm²
b = 14.05 cm

Hence, A = 50°, B =60°, C =70°, a =12.42 cm, b = 14.05 cm,c = 15.24 cm


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2.
(a)
B + D = 180°
Hence, B = 180° - D

cosB + cosD
= cos(180° - D) + cosD
= -cosD + cosD
= 0

(b)
In ΔABC (by cosine law) :
cosB = (AB² + BC² - AC²) / (2*AB*BC)
cosB = (4² + 4² - AC²) / (2*4*4)
cosB = (32 - AC²) / 32 ...... [1]

In ΔADC (by cosine law) :
cosD = (DA² + CD² - AC²) / (2*DA*CD)
cosD = (3² + 6² - AC²) / (2*3*6)
cosD = (45 - AC²) / 36 ...... [2]

[1] + [2] :
cosB + cosD = [(32 - AC²) / 32] + [(45 - AC²) / 36]
[(32 - AC²) / 32] + [(45 - AC²) / 36] = 0
36 * (32 - AC²) = -32 * (45 - AC²)]
1152 - 36AC² = -1440 + 32AC²
68AC² = 2592
AC = √(2592/68)
AC = 6.17 cm


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3.
(a)
(1/2) * PQ * RZ = A cm²
(1/2) * PQ * (20 cm) = A cm²
PQ = A/10 cm

(1/2) * QR * PX = A cm²
(1/2) * QR * (12 cm) = A cm²
QR = A/6 cm

(1/2) * RP * QY = A cm²
(1/2) * RP * (15 cm) = A cm²
PQ = 2A/15 cm

(b)
s
= (1/2)(PQ + QR + RP)
= (1/2)[(A/10) + (A/6) + (2A/15)] cm
= (1/2) * (12A/30) cm
= A/5 cm

(c)
Area of ΔPQR :
√[s * (s - PQ) * (s - QR) * (s - RP)] = A
√{(A/5) * [(A/5) - (A/10)] * [(A/5) - (A/6)] * [(A/5)- (2A/15)]} = A
√[(A/5) * (A/10) * (A/30) * (A/15)] = A
√(A⁴/22500) = A
A²/150 = A
A² - 150A = 0
A(A - 150) = 0
A = 0 (rejected) or A = 150

Area of ΔPQR = 150cm²

(d)
PQ = 150/10 cm
PQ = 15 cm

QR = 150/6 cm
QR = 25 cm

RP = 2(150)/15 cm
RP = 20 cm