Question

How do you find the values of the other five trigonometric functions of the acute angle A with `cosA=2/sqrt10`?

Answer 1

Once you have `a = 2, o = sqrt6 and h = sqrt10`

you can find `sin, cos, tan, cot, sec and "cosec"`

Explanation:

`cos A = "adj"/"hyp" = 2/sqrt10`

So, using Pythagoras:
`"opp"^2 = sqrt10^2 -2^2 = 10-4=6`

`"opp" = sqrt6`

You now have: `a = 2, o = sqrt6 and h = sqrt10`

`Sin A = o/h = sqrt6/sqrt10`

`Cos A = a/h = 2/sqrt10`

`Tan A = o/a =sqrt6/2`

`Cot A = a/o=2/sqrt6`

`Sec A = h/a =sqrt10/2`

`"cosec" A = h/o=sqrt10/sqrt6`

Answer 2

`"see explanation"`

Explanation:

`"given A is acute, that is in first quadrant then all "`
`"trig. ratioswill be positive"`

`"using "sin^2A+cos^2A=1`

`rArrsin^2A=+-sqrt(1-sin^2A)`

`•color(white)(x)cosA=2/sqrt10`

`•color(white)(x)sinA=sqrt(1-(2/sqrt10)^2)`

`color(white)(xxxxxxx)=sqrt(1-4/10)=sqrt(3/5)=sqrt3/sqrt5=sqrt15/5`

`•color(white)(x)secA=1/cosA=sqrt10/2`

`•color(white)(x)cscA=1/sinA=5/sqrt15=sqrt15/3`

`•color(white)(x)tanA=sinA/cosA=sqrt15/5xxsqrt10/2=sqrt6/2`

`•color(white)(x)cotA=1/tanA=2/sqrt6=sqrt6/3`

`"All denominators have been rationalised"`