Angles x and y are located in the first quadrant such that sinx= 3/5 and cosy= 5/13. Determine the exact value for siny and cosx

Asked on by iamelf

jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted on

Let us consider a right triangle with;

hypotenuse = h

opposite = a

From Pythagoras theorem;

`h^2 = a^2+b^2`

Then for any small angle x between b and h;

`sinx = a/h`

`cosx = b/h`

For first angle;

`a = 3`

`h = 5`

`sinx = 3/5`

`cosx = sqrt(5^2-3^2)/5 = 4/5`

For second angle;

`b = 5`

`h = 13`

`cosy = 5/13`

`siny = sqrt(13^2-5^2)/13 = 12/13 `

So the answers are;

`cosx = 4/5`

`siny = 12/13`

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

Since the problem provides the information that the angles x and y are located in the quadrant 1, hence, the values of the function `sin y`  and `cos x`  are positive.

You need to use the fundamental formula of trigonometry to evaluate cos x such that:

`sin^2 x + cos^2 x = 1`

Substituting `(3/5) ` for sin `x ` yields:

`(3/5)^2 + cos^2 x = 1 => cos^2 x = 1 - (3/5)^2`

`cos x = +-sqrt(1 - 9/25)`

Since `cos x > 0` , hence, you need to keep only the positive value,  such that:

`cos x = sqrt((25-9)/25) => cos x = sqrt(16/25)`

`cos x = 4/5`

You need to use the fundamental formula of trigonometry to evaluate `sin y`  such that:

`sin^2y + cos^2y = 1`

Substituting `(5/13)`  for `cos y`  yields:

`sin^2 y + (5/13)^2 = 1 => sin y = +-sqrt(1 - 25/169)`

Since `sin y> 0` , hence, you need to keep only the positive value, such that:

`sin y = sqrt((169-25)/169) => sin y = sqrt(144/169)`

`sin y = 12/13`

Hence, evaluating `cos x`  and `sin y` , under the given conditions, yields `cos x = 4/5`  and `sin y = 12/13` .

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