hi i hope that the answer is D
second st is easier
(2)if x=60, then angle BCD=120, and angle CBD=30, thus triangle BCD is isosceles. and BC=6.
so suff
(1) with brute force, if you know 2 sides of triangle and angle between them. you can calculate the last side using cosine theorem, no need to calculate, but simply to know
BC^2=CD^2+BD^2-2*CD*BD*cos30, now know that BD=6*3^1/2, CD=6, and cos 30 is fixed value ((3^1/2)/2)
so you can find BC
also suff
BTG quistion
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- ankurmit
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thanksclock60 wrote:hi i hope that the answer is D
second st is easier
(2)if x=60, then angle BCD=120, and angle CBD=30, thus triangle BCD is isosceles. and BC=6.
so suff
(1) with brute force, if you know 2 sides of triangle and angle between them. you can calculate the last side using cosine theorem, no need to calculate, but simply to know
BC^2=CD^2+BD^2-2*CD*BD*cos30, now know that BD=6*3^1/2, CD=6, and cos 30 is fixed value ((3^1/2)/2)
so you can find BC
also suff
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Ankur mittal
Ankur mittal


















