A rectangle has length 4 inches and width 3 inches. What is the area? 100pts
2 answers:
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Answer:
Step-by-step explanation:
tan Θ + tan 2Θ + √3 tan Θ tan 2Θ = √3
tan Θ + tan 2Θ = √3 - √3 tan Θ tan 2Θ
tan Θ + tan 2Θ = √3 ( 1 - tan Θ tan 2Θ)
(tan Θ + tan 2Θ) / (1 - tanΘ tan 2Θ) = √3
tan(Θ + 2Θ) = √3
tan 3Θ = tan () we know tan Θ = tan α; Θ = nΠ + α, n belongs to z
3Θ = nΠ + Π/3
Θ = nπ/3 + Π/9 for all n in Z
1)
∠BAC = ∠NAC - ∠NAB = 144 - 68 = 76⁰
AB = 370 m
AC = 510 m
To find BC we can use cosine law.
a² = b² + c² -2bc*cos A
|BC|² = |AC|²+|AB|² - 2|AC|*|AB|*cos(∠BAC)
|BC|² = 510²+370² - 2*510*370*cos(∠76⁰) =
|BC| ≈ 553 m
2)
To find ∠ACB, we are going to use law of sine.
sin(∠BAC)/|BC| = sin(∠ACB)/|AB|
sin(76⁰)/553 m = sin(∠ACB)/370 m
sin(∠ACB)=(370*sin(76⁰))/553 =0.6492
∠ACB = 40.48⁰≈ 40⁰
3)
∠BAC = 76⁰
∠ACB = 40⁰
∠CBA = 180-(76+40) = 64⁰
Bearing C from B =360⁰- 64⁰-(180-68) = 184⁰
4)
Shortest distance from A to BC is height (h) from A to BC.
We know that area of the triangle
A= (1/2)|AB|*|AC|* sin(∠BAC) =(1/2)*370*510*sin(76⁰).
Also, area the same triangle
A= (1/2)|BC|*h = (1/2)*553*h.
So, we can write
(1/2)*370*510*sin(76⁰) =(1/2)*553*h
370*510*sin(76⁰) =553*h
h= 370*510*sin(76⁰) / 553= 331 m
h=331 m
