How do I draw the right triangle ABC whose sides have the following values given below, and then find the six trigonometric functions of each triangle’s angle A? a = 2, b= #sqrt5# , c = 3

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Answer 1 · 2018-07-23T11:57:52

# sin theta = b/c=sqrt5/3, cos theta = a/c=2/3 , tan theta= b/a=sqrt 5/2 , csc theta = c/b=3/sqrt 5 , sec theta= c/a=3/2 ,cot theta= a/b=2/sqrt 5#

#a= 2 :. a^2=4 , b=sqrt 5 :. b^2=5 , c=3 :. c^2=9#

#a^2+b^2= 5+4=9=c^2:. a^2+b^2= c^2 ; a,b,c # are

adjacent side , opposite side and hypotenuse of the right

triangle. To be drawn making sides #a and b# at right angle

and #c# as hypotenuse of the triangle.

Let #theta# be the angle between adjacent side #a#

and hypotenuse #c# , then , # sin theta = b/c=sqrt5/3#

# cos theta = a/c=2/3 , tan theta= b/a=sqrt 5/2# ,

# csc theta = c/b=3/sqrt 5 , sec theta= c/a=3/2# ,

#cot theta= a/b=2/sqrt 5# [Ans]