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Grade: 8

                        

If in triangle ABC with right angle C, sinA=3/5,then what is the value of sinB?

2 years ago

Answers : (2)

Susmita
425 Points
							
Please draw a right angle triangle where angle ACB is 90°.
Given
sinA=sin\angle CAB =\frac{BC}{AB}=\frac{3}{5}
Let BC=3 and AB=5.Using pythagoras law
AC2=AB2-BC2=25-9=16
Or,AC=4
So sinB=sin\angle ABC=\frac{AC}{AB}=\frac{4}{5}
Hope this helps
2 years ago
Soumendu Majumdar
159 Points
							
{\color{Red} sin A = 3/5}
We know that sum of all internal angles in a triangle = 180^{\circ}}
 
\therefore \angle A +\angle B+\angle C=180^{\circ}
\therefore \angle A +\angle B=90^{\circ}
\therefore \angle B =90^{\circ}-\angle A
sin(90^{\circ}-\angle A)=cos A
\Rightarrow sin B = cos A = \sqrt{1-sin^{2}A}=\sqrt{1-(9/25)}=\sqrt{16/25}=4/5
Hope it helps!
If you have any query feel free to ask we’ll be glad to help you!
one year ago
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