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

Let us  calculate the length of AB 

In Δ ABC, using Pythagoras theorem,

$AC^2 = AB^2 + BC^2$

Hence AB =

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

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

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

We have 

n2−nC2=n2−nC10

We know that

nCx=nCy⇒x=y or x+y=n

⇒2+10=n2−n

⇒n2−n=12

⇒n2−n−12=0

⇒n2−4n+3n−12

⇒n(n−4)+3(n−4)=0

⇒(n+3)(n−4)=0

⇒n=4, −3

So, n=4  [n is non negative integer]