The lengths of the sides of a triangle are consecutive integers and the largest angle is twice the smallest angle. find the measure of the smallest angle to the nearest degree.

  41°

Step-by-step explanation:

I could not think of an easy way to solve this, apart from having a graphing calculator do it. In the end, I found I could solve it analytically using a combination of the law of sines and the law of cosines.

Let x represent the length of the shortest side, and θ the smallest angle. Then the law of sines tells you ...

  sin(θ)/x = sin(2θ)/(x+2)

Cross-multiplying and using the trig identity for sin(2θ), we have ...

  (x +2)sin(θ) = 2x·sin(θ)cos(θ)

Dividing out sin(θ), we see that ...

  cos(θ) = (x+2)/(2x)

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The law of cosines for the shortest side and smallest angle tells you ...

  x^2 = (x+1)^2 + (x+2)^2 - 2(x+1)(x+2)·cos(θ)

Substituting the above expression for cos(θ), this can be rewritten as ...

  0 = (x^2 +2x +1) +(x^2 +4x +4) -x^2 -(x+1)(x+2)^2/x

  0 = x^2 +6x +5 -(x+1)(x+2)^2/x . . . . . . collect terms outside the fraction

  0 = x(x+5)(x+1) -(x+1)(x+2)^2 . . . . . . . . factor and multiply by x

We know that x=-1 is not a solution, so we can divide by that factor:

  0 = x^2 +5x -(x^2 +4x +4) . . . . . multiply it all out

  0 = x -4 . . . . . . . . . . . . . . . . . . . . collect terms

  4 = x

so, cos(θ) = (4+2)/(2·4) = 6/8 = 3/4

and the angle of interest is ...

  θ = arccos(3/4) ≈ 41.40962° ≈ 41°

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The attachment shows a triangle-solver's result using the consecutive integers for side lengths. It confirms the answer we have here.

45

step-by-step explanation: