Question

Consider right triangle ABC, right angled at B. If AC = 17 units and BC = 8 units determine all the trigonometric ratios of angle C.​

Answer 1

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Let us  calculate the length of AB 

In Δ ABC, using Pythagoras theorem,

AC^2 = AB^2 + BC^2

Hence AB =

$\sqrt{ {ac}^{2} - {bc}^{2} }$

$\sqrt{ {17}^{2} - {8}^{2} } = \sqrt{225} = 15units$

AB = 15 units .

 

Now let us calculate the trigonometric ratios of angle C

sin C = opposite side/hypotenuse=AB/AC= 15/17

 

cos C = adjcent side/hypotenuse=BC/AC= 8/17

 

tan C =opposite side/adjcent side =AB/BC= 15/8

 

cot C =1/tan C =adjcent side/opposite side = BC/AB = 8/15

 

sec C = 1/cos C = hypotenuse/adjcent side = AC/BC= 17/8

 

cosec C = 1/sin C = hypotenus/eopposite side =AC/AB= 17/15

Answer 2

Step-by-step explanation:

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Refer to the above attachment ⬆️⬆️.

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