Show that the area of Δ快! 40分!

Q1Method 1 :(a² sin B sin C) / {2 sin (B+C)}= (a² sin B sin C) / (2 sin A)= (1/2) (a² sin C) (sinB / sinA)= (1/2) (a² sin C) (b/a)= (1/2) (a b sinC)= △ABC
Method 2 :Note that sin(B+C) = sin(180° - A) = sin A = h/bsinB = h/aSo (a² sin B sin C) / {2 sin (B+C)}= (a² sin B sin C) / (2 sin A)= (a² h/a sin C) / (2 h/b)= (1/2) (a b sinC)= △ABC
Q2a)ㄥPOR = θ + ψ ,Cosine formula :L² = 10² + 5² - 2(10)(5) cos(θ + ψ)L² = 125 - 100 cos(θ + ψ)L = 5 √( 5 - 4 cos(θ + ψ) )
bi)11 = 5 √( 5 - 4 cos(θ + 45°) )cos(θ + 45°) = 0.04θ + 45° = 87.7°θ = 42.7°
bii)By the result of part a) :L' = 5 √( 5 - 4 cos(2θ + ψ) )L' = 5 √( 5 - 4 cos(2*42.7° + 45°) )L' = 13.78
The difference between L and L' = 13.78 - 11 = 2.78 km

Question 1.( a²sin B sin C)/ {2 sin (B+C)}( a sin B a sin C) / {2 sin (180° - A)}Since sin B = h/a, h= a sin B( h a sin C) / {2 (sin 180° cos A - cos 180° sinA)} ( h a sin C) / {2 ((0) cos A – (-1) sinA)} ( h a sin C) / {2 sinA)} (h/2)(a sin C/sin A) ----------- (1) Using sine law, a/sinA = c/sinC, a = a sin C /sin ASubstitute c = a sin C /sin A into equation (1)(h/2)(c)(h)(c)/2 is equalto the area of triangle ABC where c is base, h is height
Question 2:(a) Consider triangle PQR, angle PQR = q + f Using cosine law, L^2 = 10^2 + 5^2 – 2(10)(5) cos (q + f)L = square root of (10^2 + 5^2 – 2(10)(5) cos (q + f))L = square root of (125 – 100 cos (q + f)) (b) L = square root of (125 – 100 cos (q + f))11 = square rootof (125 – 100 cos (q+ 45°))Take square on both sides121 = 125 – 100 cos (q + 45°)100 cos (q + 45°) = 125 – 121 = 4cos (q + 45°) = 4/100 = 0.04(q+ 45°)= cos^-1(0.04)q+ 45°= 87.7°q = 87.7° - 45°q = 42.7° (c) L’ = square root of (125 – 100 cos (2 * 42.7° + 45°)) L’ = square rootof (125 – 100 cos (130.4°)) L’ = square rootof (125 – 100 (-0.6481)) L’ = square rootof (125 + 64.8120) L’ = 13.777 km L’ – L = 13.777km – 11 km = 2.777 kmDifference between L and L’ is 2.777 km