Maths,θ - Ans.fyi 參考答案

Maths,θ

2010-01-07 5:21 am

show step,answer in 2d.p

1.Given 三角形ABC as shown,AB=7cm ,BC=8cm, CA=9cm ,find
a. angles A,B and C;
b. area of 三角形ABC

2, Find &theta; where 0 &deg; &lt; / = &theta; &lt; / = 180 &deg;
a. cos &theta; =0.5
b. 3sin2 &theta; =1.2
c. 2tan^2 &theta; - 9tan &theta; =5

3.a) Show tan(180+ &theta; )=tan &theta;
b) Show cos3A= 4cos^3A - 3cosA.

更新1:

2. Find 0 where 0度小於或等於 0 小於或等於 180度 a) cos 0 =0.5 b) 3sin2 0 =1.2 c) 2tan^2 0 - 9tan 0 =5

回答 (1)

用戶9112

2010-01-07 5:48 am

✔ 最佳答案

(1)(a) cosA
= (b^2 + c^2 - a^2)/2bc
= (81 + 49 - 64)/(2*7*9)
= 0.5238
A = 58.41 degrees
cosB
= (a^2 + c^2 - b^2)/(2*a*c)
= (64 + 49 - 81)/(2*7*8)
= 0.2857
B = 73.40 degrees
C = 180 - A - B = 48.19 degrees
(b) Area = (1/2)(AB)(AC)sinA
= (1/2)(7)(9)sin58.41
= 26.83 cm^2
(2) (a) cosθ = 0.5
cosθ = cos60
θ = 60
(b) 3sin2θ = 1.2
sin2θ = 0.4
sin2θ = sin23.58
2θ = 180n + (-1)^n*23.58
θ = 90n + (-1)^n*11.79
θ = 11.79 or 78.21
(c) 2tan^2θ - 9tanθ = 5
2tan^2θ - 9tanθ - 5 = 0
(2tanθ + 1)(tanθ - 5) = 0
tanθ = -1/2 or tanθ = 5
θ = 153.43 or θ = 78.69
(3)(a) tan(180+ θ)
= (tan180 + tanθ)/(1 - tan180tanθ)
= (0 + tanθ)/(1-0*tanθ)
= tanθ
b) cos3A
=cos(2A+A)
= cos2AcosA - sin2AsinA
= (cosAcosA - sinAsinA)cosA - 2sinAcosAsinA
= (2cos^2A - 1)cosA - 2cosA(1 - cos^2A)
= 2cos^3A - cosA - 2cosA + 2cos^3A
= 4cos^3A - 3cosA

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