In the triangle ABC, if the angle a = 60 degrees and BC = 3, then the perimeter of the triangle ABC is?

From the sine theorem, we have
BC/sinA=AC/sinB=AB/sinC
AC = bcsinb / Sina = 3sinb / sin60 ° = 3sinb / (√ 3 / 2) = 2 √ 3sinb
AB=BCsinC/sinA=BCsin[180°-(A+B)]/sinA=3sin(60°+B)/sin60°
=(3sin60°cosB+3sinBcos60°)/sin60°
=3cosB+3sinBcot60°
=3cosB+√3sinB
AB+BC+AC=3cosB+√3sinB+3+2√3sinB=3√3sinB+3cosB+3
The perimeter of Δ ABC is 3 √ 3sinb + 3cosb + 3

In triangle ABC, angle a = 60 degrees, BC = 3, find the perimeter of triangle ABC

If it's a right triangle, it's simpler to use trigonometric function directly; if it's a general triangle, you need to know one corner on both sides before you can use cosine theorem to find the other side and then the perimeter; if it's an equilateral triangle, it's the simplest, three sides are 3 and the perimeter is 9

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