bantu ganJika a dan ß adalah sudut-sudut yang lancip, sin(a -B) = 1/2√3, dan sin a cos ß = 1/2, maka.... Question from @Rynzn

Jawaban:

$4 \sqrt{3} + 6$

Penjelasan dengan langkah-langkah:

sin (a-b)= sin a cos b - sin b cos a.

$\frac{1}{2} \sqrt{3} = \frac{1}{2} - \sin( \beta ) \cos( \alpha ) \\ \frac{1}{2}( \sqrt{3} - 1) = - \sin( \beta ) \cos( \alpha ) \\ - \frac{1}{2} ( \sqrt{3} - 1) = \sin( \beta ) \cos( \alpha )$

$\frac{ \sin( \alpha + \beta ) }{ \sin( \alpha - \beta ) } \\ = \frac{ \sin( \alpha ) \cos( \beta ) + \sin( \beta ) \cos( \alpha ) }{ \sin( \alpha ) \cos( \beta ) - \sin( \beta ) \cos( \alpha ) } \\ = \frac{ \frac{1}{2} + \frac{1}{2}( \sqrt{3} - 1)}{ \frac{1}{2} - \frac{1}{2} ( \sqrt{3} - 1)} \\ = \frac{ \sqrt{3} }{1 - \frac{1}{2} \sqrt{3} } \\ = \frac{ \sqrt{3}(1 + \frac{1}{2} \sqrt{3}) }{(1 - \frac{1}{2} \sqrt{3})(1 + \frac{1}{2} \sqrt{3} ) } \\ = \frac{ \sqrt{3} + \frac{3}{2} }{1 - \frac{3}{4} } \\ = \frac{ \sqrt{3} + \frac{3}{2} }{ \frac{1}{4} } \\ = 4 \sqrt{3} + 6$

4 votes Thanks 10

Fazm1

sin (a - b) = sin a cos b - sin b cos a

½√3 = ½ - sin (β) cos (α)

½(√3 - 1) = - sin (β) cos (α)

-½(√3 - 1) = sin (β) cos (α)

sin (α + β) / sin (α - β)

= sin (α) cos (β) + sin (β) cos (α)

sin (α) cos (β) - sin (β) cos (α)

= ½ + ½ (√3 - 1) / ½ - ½ (√3 - 1)

= √3 / 1 - ½√3

= √3 (1 + ½√3) / (1 - ½√3)(1 + ½√3)

= √3 + 3/2 / 1 - ¾

= √3 + 3/2 / ¼

= 4√3 + 6

2 votes Thanks 8

Rynzn 1/2(√3-1) darimna bang?

kelvinho018527 pindah ruas ke kanan, faktorin 1/2

kelvinho018527 *kiri

vquinn1688 udah dijawab bukannya makasih

Rynzn ohh...iyaa makasih bang @kelvinho018527 lupa gk ada kouta

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