The sides of a triangle are in ratio 2:3:4. Find the cosine of the largest angle
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# The sides of a triangle are in ratio 2:3:4. Find the cosine of the largest angle

[From: ] [author: ] [Date: 12-12-20] [Hit: ]
difficult part is determining the largest angle. Easiest is to do all three.A = arccos 3/12 = 104.B = arccos 21/24 = 29.C = arccos 11/16 = 46.so the answer is –0.......
cosine rule relates the lengths of the sides of a
plane triangle to the cosine of one of its angles.
If a, b, c are the three sides of a triangle, and
C is the angle between a and b and opposite side c,
then:
c² = a² + b² – 2abcosC
or
cos C = (a² + b² – c²) / (2ab)

difficult part is determining the largest angle. Easiest is to do all three.

let angle A be between sides 2 and 3
cos A = (2² + 3² – 4²) / (2•2•3) = –3/12
let angle B be between sides 3 and 4
cos B = (3² + 4² – 2²) / (2•3•4) = 21/24
let angle C be between sides 2 and 4
cos C = (2² + 4² – 3²) / (2•2•4) = 11/16
A = arccos 3/12 = 104.5º
B = arccos 21/24 = 29.0º
C = arccos 11/16 = 46.6º

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Let:
x = perimeter of the triangle
cos theta = cosine of the largest angle

longest side = 4x/(2+3+4 = 4x/9

one side = 2x/9
another side = 3x/9

Using cosine law, the cosine theta of the largest side is:

(4x/9)^2 = (2x/9)^2 + (3x/9)^2 - (2)(2x/9)(3x/9)(cos theta)
(16x^2)/81 = (4x^2)/81 + (9x^2)/81 - (cos theta * 12x^2)/81
(cos theta * 12x^2)/81 = (-3x^2)/81
cos theta * 12 = -3
cos theta = -3/12
cos theta = -0.25
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