In ΔABC right angled at B, sin C = 513. Find Cos A.

We can say sin C = side opposite to ∠Chypotenuse=ABAC= 513So let AB = 5k and AC = 13kBy using Pythagoras theorem, BC = AC2−AB2So BC = (13k)2−(5k)2 = 169k2−25k2 = 144k2 = 12k∴cos A = side adjcent to ∠ Ahypotenuse = ABAC= 5k13k=513.

In ΔABC right angled at B, sin C = 513. Find Cos A.

We can say sin C = side opposite to ∠Chypotenuse=ABAC= 513So let AB = 5k and AC = 13kBy using Pythagoras theorem, BC = AC2−AB2So BC = (13k)2−(5k)2 = 169k2−25k2 = 144k2 = 12k∴cos A = side adjcent to ∠ Ahypotenuse = ABAC= 5k13k=513.

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In ΔABC right angled at B, sin C = $\frac{5}{13}$ . Find Cos A.

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$\frac{5}{13}$

$\frac{12}{13}$

$\frac{4}{13}$

None

answer is A.

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Detailed Solution

We can say sin C = $\frac{side opposite to ∠C}{hypotenuse}$ = $\frac{AB}{AC}$ = $\frac{5}{13}$
So let AB = 5k and AC = 13k

By using Pythagoras theorem, BC = $\sqrt{A C^{2} - A B^{2}}$
So BC = $\sqrt{(13 k)^{2} - (5 k)^{2}}$ = $\sqrt{169 k^{2} - 25 k^{2}}$ = $\sqrt{144 k^{2}}$ = 12k