# Thread: Question help (from cambridge 3 unit year 11)

1. ## Question help (from cambridge 3 unit year 11)

Hi,

I came a cross a question from exercise 8D and i'm utterly confused.

7. a) If ABCD is a parallelogram, show that SinA = SinB = SinC = SinD.

It said that it is related to trigonometry and incorporated trigonometric work. Does it involve the supplement of the parallelogram or have anything to do with the sin angle?

2. ## Re: Question help (from cambridge 3 unit year 11)

Originally Posted by dabatman
Hi,

I came a cross a question from exercise 8D and i'm utterly confused.

7. a) If ABCD is a parallelogram, show that SinA = SinB = SinC = SinD.

It said that it is related to trigonometry and incorporated trigonometric work. Does it involve the supplement of the parallelogram or have anything to do with the sin angle?

$\noindent Well, \sin{A} = \sin{C} as \angle A = \angle C, being opposite angles in a parallelogram. And \sin{A} = \sin{(180-A)}. But 180 - A = \angle B = \angle D, being co-interior angles; supplementary. Thus, \sin{(180 - A)} = \sin{A} = \sin{B} = \sin{C} = \sin{D}, as required.$

3. ## Re: Question help (from cambridge 3 unit year 11)

Originally Posted by 1729
$\noindent Well, \sin{A} = \sin{C} as \angle A = \angle C, being opposite angles in a parallelogram. And \sin{A} = \sin{(180-A)}. But 180 - A = \angle B = \angle D, being co-interior angles; supplementary. Thus, \sin{(180 - A)} = \sin{A} = \sin{B} = \sin{C} = \sin{D}, as required.$
thank you so much

4. ## Re: Question help (from cambridge 3 unit year 11)

+1

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