Given that ∆ABC is an equilateral triangle of side 1cm , ∆BDC is iscoceles triangle with DB = DC outward of ∆ABC and angle BDC =120 . If points M and N are on AB and AC respectively such that angle MDN =60. find perimeter of ∆AMN
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Step-by-step explanation:
Given, ABC is an equilateral
BDC is an isosceles right triangle
angle CBD = angle BCD = x ( angles of isosceles triangle)
In triangle BDC
x + x + angle D = 180 ( angle sum property)
2x = 180 - 90
x = 90/2
x = 45
IN triangle ABC
angle ABC = 60 ( angles of an equilateral are equal to 60)
angle ABD = x + angle ABC
Angle ABD = 45+60
Angle ABD = 105
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