Let The Angles Of A Triangle Be α, β, And γ, Wi

Anonymous

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Let the angles of a triangle be α, β, and γ, with opposite sides of length a, b, and c, respectively. Use the Law of Cosines and the Law of Sines to find the remaining parts of the triangle. (Round your answers to one decimal place.) a = 4; b = 5; c = 7 α = β = γ =

 

Answer No.1

CrazyTeacher9

From: US
Posts: 22
Votes: 88
C = a + b - 2ab cos(?) b = a + c - 2ac cos() a = b + c - 2bc cos(a) Solving these give: ? = arccos[(a + b - c)/(2ab)] = 120 = arccos[(a - b + c)/(2ac)] = 38.21 a = arccos[(-a + b + c)/(2bc)] = 21.79 The sum 120 + 38.21 + 21.79 = 180, as it should. To find the area, we need to use one side as the base and find the height to the third vertex. Letting side c be our base, we need to find the height. Thinking of the triangle as two right triangles side-by-side, we need to find the base length of each. The sum of the base lengths is 7, so call one base b and call the other (7-b). 4 = h + b 5 = h + (7-b) Solve the system of equations: h = sqrt(16 - b) h = sqrt(25 - (7-b)) So, sqrt(16 - b) = sqrt(25 - 49 + 14b - b) = sqrt(-24 + 14b - b) 16 - b = -24 + 14b - b 16 = -24 + 14b 14b = 40 b = 40/14=20/7 h = sqrt(16 - 20/7) =2.799 Now use the formula for area of a triangle: 0.5bh = 0.5*7*1.8558 = 6.4953 square units

Answer No.2

WizNewton

From: US
Posts: 26
Votes: 52
A = 4; b = 5; c = 7 7^2 = 4^2 + 5^2 - 2*4*5 cosc 49 = 16+25 - 40 cos c 8 = -40 cosc cosc = -8/40 c = cos^-1(-1/5) = 101.53695903281392 a/sinA = c/SinC sin A = a*sin(C)/c = 4/7 sin(101.53695903281392) = 0.55988336977901714286 A = 34.04 α = 34.04 β = 180 - 34.04 - 101.53 = 44.42 γ =101.53695903281392

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