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

Rayan , 6 Years ago

Grade 8

2 Answers

Susmita

Last Activity: 6 Years ago

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

Soumendu Majumdar

Last Activity: 6 Years ago

${\color{Red} sin A = 3/5}$

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!

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