In △ABC, ∠B=90∘, sinC=53, then cosA= ______

In a right angled triangle, A+B+C=180∘B=90∘So, A+C=90∘A=90∘−CcosA=cos(90∘−A)⇒cosA=sinC=53⇒cosA=53

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In ABC, B= {90}^{}, {C} = {3}{5}, then {A} = ______

Question

In $△ A BC$ , $∠ B = 90^{\circ}$ , $sin C = \frac{3}{5}$ , then $cos A =$ ______

\frac{3}{5}

\frac{4}{5}

\frac{5}{4}

\frac{5}{3}

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Owen Discussed

△ A BC

∠ B = 90^{\circ}

sin C = \frac{3}{5}

, then

cos A =

______

10 mins ago

Discuss this question LIVE

10 mins ago

In a right angled triangle,

A + B + C = 18 0^{\circ}

B = 9 0^{\circ}

So,

A + C = 9 0^{\circ}

A = 9 0^{\circ} - C

cos A = cos (9 0^{\circ} - A)

\Rightarrow cos A = sin C = \frac{3}{5}

\Rightarrow cos A = \frac{3}{5}

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Question Text	In $△ A BC$ , $∠ B = 90^{\circ}$ , $sin C = \frac{3}{5}$ , then $cos A =$ ______
Answer Type	Text solution:1
Upvotes	150