Question

1 + Sin 60+ cost 60 by
sign 30+ cost 30

Answer 1

SOLUTION:

$\\ \\ = \frac{1 + \sin(60) + \cos(60) }{ \sin(30) + \cos(30) } \\ \\ = \frac{1 + \frac{ \sqrt{3} }{2} + \frac{1}{2} }{ \frac{1}{2} + \frac{ \sqrt{3} }{2} } \\ \\ = (\frac{2 + \sqrt{3} \: + 1}{2} ) \div ( \frac{1 + \sqrt{3} }{2} ) \\ \\ = \frac{3 + \sqrt{3} }{1 + \sqrt{3} } \\ \\ = \frac{(3 + \sqrt{3}) \: (1 + \sqrt{3} )}{ {(1)}^{2} - ( { \sqrt{3} })^{2} } \\ \\ = \frac{3(1 + \sqrt{3} ) \sqrt{3} (1 + \sqrt{3} )}{2} \\ \\ = \frac{3 + 3\sqrt{3} + \sqrt{3} + 3 }{2} \\ \\ = \frac{6 + 4 \sqrt{3} }{2} \\ \\ = \frac{2(3 + 2\sqrt{3}) }{2} \\ \\ = 3 + 2 \sqrt{3} \: \: \: \: answer$

Answer 2

Answer:

∴ ∠ABC = 90° [angle in semi circle] ---- (i)

Also, AD is a diameter of the circle with center O .

∴ ∠ABD = 90° [angle in semi circle] ---- (ii)

on adding Eqns, (i) and (ii) we get

⇒ ∠ABC + ∠ABD = 180°

So. CBD is a straight line.

Hence C, B and D are collinear . Hence proved.