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
AC^{2} = AB^{2} + BC^{2}
5^{2} = 4^{2} + BC^{2}
25 = 16 + BC^{2}
25 - 16 = BC^{2}
9 = BC^{2}
BC = √9
BC = 3
Problem 2 :
Solution :
Using reference triangle.
Here Opposite = 7, Adjacent = 24
AC^{2} = AB^{2} + BC^{2}
= 7^{2} + 24^{2}
= 49 + 576
AC^{2} = 625
AC = √625
AC = 25
Problem 3 :
Solution :
Using reference triangle.
Here Adjacent = 5, Hypotenuse = 13
AC^{2} = AB^{2} + BC^{2}
13^{2} = AB^{2} + 5^{2}
169 = AB^{2} + 25
169 - 25 = AB^{2}
144 = AB^{2}
AB = √144
AB = 12
Problem 4 :
Solution :
Using reference triangle.
Here Opposite = 5, Hypotenuse = 13
AC^{2} = AB^{2} + BC^{2}
13^{2} = 5^{2} + BC^{2}
169 = 25 + BC^{2}
169 - 25 = BC^{2}
BC^{2 } = 144
BC = √144
BC = 12
Problem 5 :
Solution :
Using reference triangle.
Here Opposite = = -3, Hypotenuse = 5
AC^{2} = AB^{2} + BC^{2}
5^{2} = (-3)^{2} + BC^{2}
25 = 9 + BC^{2}
BC^{2} = 25 - 9
BC^{2} = 16
BC = √16
BC = 4
Problem 6 :
Solution :
Using reference triangle.
Here Opposite = = -4, Hypotenuse = 5
AC^{2} = AB^{2} + BC^{2}
5^{2} = (-4)^{2} + BC^{2}
25 = 16 + BC^{2}
BC^{2} = 16 - 25
BC^{2} = -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
AC^{2} = AB^{2} + BC^{2}
= (2√2)^{2} + 1^{2}
= 8 + 1
= 9
AC^{2} = 9
AC = √9
AC = 3
Problem 9 :
Solution :
Using reference triangle.
Here Hypotenuse = 7, Adjacent = 6
AC^{2} = AB^{2} + BC^{2}
7^{2} = AB^{2} + 6^{2}
49 = AB^{2} + 36
49 - 36 = AB^{2}
13 = AB^{2}
AB = √13
Problem 10 :
Solution :
Using reference triangle.
Here Adjacent = √21, Hypotenuse = 7
AC^{2} = AB^{2} + BC^{2}
7^{2} = AB^{2} + (√21)^{2}
49 = AB^{2} + 21
49 - 21 = AB^{2}
28 = AB^{2}
AB = √28
AB = 2√7
Problem 11 :
Solution :
Using reference triangle.
Here Opposite = -√11, Adjacent = 3
AC^{2} = AB^{2} + BC^{2}
= (-√11)^{2} + 3^{2}
= 11 + 9
= 20
AC = √20
AC = 2√5
Problem 12 :
Solution :
Here Hypotenuse = 7, Adjacent = 2
AC^{2} = AB^{2} + BC^{2}
7^{2} = AB^{2} + 2^{2}
49 = AB^{2} + 4
49 - 4 = AB^{2}
45 = AB^{2}
AB = √45
AB = 3√5
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM