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


2 years ago

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 lawAC2=AB2-BC2=25-9=16Or,AC=4So $sinB=sin\angle ABC=\frac{AC}{AB}=\frac{4}{5}$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!

2 years ago
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### Course Features

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### Course Features

• 31 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions