In a $Δ A B C, ∠ B = 90^{o}, A B = 12 c m$ and BC = 5 cm.

Find (i) cos A (ii) cosec A (iii) cos C (iv) cosec C.

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Solution

By Pythagoras theorem
$A C^{2} = A B^{2} + B C^{2}$
$A C = \sqrt (A B^{2} + B C^{2})$
$A C = \sqrt (12^{2} + 5^{2})$
$A C = \sqrt 144 + 25 = \sqrt 169 = 13$

(i) cos A=AB/AC=12/13

(ii) cosec A =AC/BC=13/5

(iii) cos C=BC/AC=5/13

(iv) cosec C=AC/AB=13/12

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In a ΔABC, ∠B=90o,AB=12 cm and BC = 5 cm. Find (i) cos A (ii) cosec A (iii) cos C (iv) cosec C.

By Pythagoras theorem AC2=AB2+BC2 AC=√(AB2+BC2) AC=√(122+52) AC=√144+25=√169=13 (i) cos A=AB/AC=12/13 (ii) cosec A =AC/BC=13/5 (iii) cos C=BC/AC=5/13 (iv) cosec C=AC/AB=13/12

In a $Δ A B C, ∠ B = 90^{o}, A B = 12 c m$ and BC = 5 cm.

Find (i) cos A (ii) cosec A (iii) cos C (iv) cosec C.

By Pythagoras theorem
$A C^{2} = A B^{2} + B C^{2}$
$A C = \sqrt (A B^{2} + B C^{2})$
$A C = \sqrt (12^{2} + 5^{2})$
$A C = \sqrt 144 + 25 = \sqrt 169 = 13$

(i) cos A=AB/AC=12/13

(ii) cosec A =AC/BC=13/5

(iii) cos C=BC/AC=5/13

(iv) cosec C=AC/AB=13/12