To evaluate composite inverse trig functions, we should aware of reference triangle.
A reference triangle is formed by "dropping" a perpendicular from the terminal ray of a standard position angle to the x-axis. Remember, it must be drawn to the x-axis
With the help of formulas given above, we will rewrite the given inverse trig functions in terms of other inverse trig functions.
Find the exact value of each expression.
Problem 1 :
Solution :
Using reference triangle.
Here, Opposite = 4, Hypotenuse = 5
AC2 = AB2 + BC2
52 = 42 + BC2
25 = 16 + BC2
25 - 16 = BC2
9 = BC2
BC = √9
BC = 3
Problem 2 :
Solution :
Using reference triangle.
Here Opposite = 7, Adjacent = 24
AC2 = AB2 + BC2
= 72 + 242
= 49 + 576
AC2 = 625
AC = √625
AC = 25
Problem 3 :
Solution :
Using reference triangle.
Here Adjacent = 5, Hypotenuse = 13
AC2 = AB2 + BC2
132 = AB2 + 52
169 = AB2 + 25
169 - 25 = AB2
144 = AB2
AB = √144
AB = 12
Problem 4 :
Solution :
Using reference triangle.
Here Opposite = 5, Hypotenuse = 13
AC2 = AB2 + BC2
132 = 52 + BC2
169 = 25 + BC2
169 - 25 = BC2
BC2 = 144
BC = √144
BC = 12
Problem 5 :
Solution :
Using reference triangle.
Here Opposite = = -3, Hypotenuse = 5
AC2 = AB2 + BC2
52 = (-3)2 + BC2
25 = 9 + BC2
BC2 = 25 - 9
BC2 = 16
BC = √16
BC = 4
Problem 6 :
Solution :
Using reference triangle.
Here Opposite = = -4, Hypotenuse = 5
AC2 = AB2 + BC2
52 = (-4)2 + BC2
25 = 16 + BC2
BC2 = 16 - 25
BC2 = -9
BC = √-9
BC = 3
Problem 7 :
Solution :
Using reference triangle.
Here Opposite = √7, Adjacent = 3
Problem 8 :
Solution :
Using reference triangle.
Here Opposite = 2√2, Adjacent = 1
AC2 = AB2 + BC2
= (2√2)2 + 12
= 8 + 1
= 9
AC2 = 9
AC = √9
AC = 3
Problem 9 :
Solution :
Using reference triangle.
Here Hypotenuse = 7, Adjacent = 6
AC2 = AB2 + BC2
72 = AB2 + 62
49 = AB2 + 36
49 - 36 = AB2
13 = AB2
AB = √13
Problem 10 :
Solution :
Using reference triangle.
Here Adjacent = √21, Hypotenuse = 7
AC2 = AB2 + BC2
72 = AB2 + (√21)2
49 = AB2 + 21
49 - 21 = AB2
28 = AB2
AB = √28
AB = 2√7
Problem 11 :
Solution :
Using reference triangle.
Here Opposite = -√11, Adjacent = 3
AC2 = AB2 + BC2
= (-√11)2 + 32
= 11 + 9
= 20
AC = √20
AC = 2√5
Problem 12 :
Solution :
Here Hypotenuse = 7, Adjacent = 2
AC2 = AB2 + BC2
72 = AB2 + 22
49 = AB2 + 4
49 - 4 = AB2
45 = AB2
AB = √45
AB = 3√5
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM