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

2 Answers
Oct 28, 2017

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#

Oct 28, 2017

#"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"#