Question

If secA = 13÷12, calculate sinA and CosA (A is an acute angle)
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Answer 1

Answer:

Let △ABC be right angled triangle (right angled at B)

secQ= AB/AC

= 12/13

Let AC=13x and AB=12x

AC²=AB²+BC²

BC² =AC² −AB²

x² =169−144

x² =25

x= ✓ 25

=5

sinA=5/13

sinA=5/13cosA=12/13