In fig. triangle ABC, AD is perpendicular to BC, Angle B = 60, Angle C = 30, AD = 2root3, find BC
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2 thoughts on “In fig. triangle ABC, AD is perpendicular to BC, Angle B = 60, Angle C = 30, AD = 2root3, find BC<br />​”

1. Step-by-step explanation:

   In triangle ABC , A=30° , let b=2k then c=√3k.

   cosA= (b^2+c^2-a^2)/2.b.c.

   cos 30° = (4k^2+3k^2-a^2)/(2.2k.√3.k)

   √3/2 = (7k^2-a^2)/4√3.k^2.

   or , √3/1= (7k^2-a^2)/(2√3k^2)

   or , 6k^2 = 7k^2 -a^2.

   or , a^2 = k^2.

   or , a= k………………………….(1)

   cosB = (c^2+a^2-b^2)/(2c.a) = (3k^2+k^2-4k^2)/(2.√3.k.k)

   cos B = 0 => B =90°.

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2. Step-by-step explanation:

   -Oct-2019 · 1 answer

   In ABC, if AD is perpendicular to BC and BD=10cm, angleB=60° and angle C =30°, then find CD.. Get the answers you …

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   20-Jun-2019 · 15 answers

   Since ad is a perpendicular bisector the angle 90 deg at BAC gets splitted in the ratio 1:2. Since the angles in them are 30 deg and 60 deg

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