Question help (from cambridge 3 unit year 11) (1 Viewer)

dabatman · Mar 5, 2017

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?

Please help

1729 · Mar 5, 2017

dabatman said:
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?

Please help

$$\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.$

dabatman · Mar 5, 2017

1729 said:
$$\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

dabatman · Mar 5, 2017

Question help (from cambridge 3 unit year 11) (1 Viewer)

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