# EVALUATING COMPOSITE INVERSE TRIG FUNCTIONS

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

BC = 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

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