Question

1) The angles of a triangle are 105 and 15°, find the ratio of its sides.​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

a:b:c= $\frac{ (\sqrt{6} +\sqrt{2} )}{4}: \frac{\sqrt{3}-1 }{2\sqrt{2} }: \frac{\sqrt{3} }{2}$

In triangle ABC,

Where ∠ A= 105°,∠B = 15°,∠C= 180°- (105°+15°)

∠C= 180°-120

  ∠C= 60°

By the sine rule,

 $\frac{a}{sin A} =\frac{b}{sin B} =\frac{c}{sin C}\\\frac{a}{sin105 } =\frac{b}{sin 15,} =\frac{c}{sin 60}\\\frac{a}{sin105 }=k (Let k be the constant) \\a= k ( sin105)=k\frac{ (\sqrt{6} +\sqrt{2} )}{4}\\b=k ( sin15)= k\frac{\sqrt{3}-1 }{2\sqrt{2} }\\c= k ( sin60)=k\frac{\sqrt{3} }{2}$                                        

Thus,

a:b:c= $\frac{ (\sqrt{6} +\sqrt{2} )}{4}: \frac{\sqrt{3}-1 }{2\sqrt{2} }: \frac{\sqrt{3} }{2}$

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