Answer :
Let's go through each part of the question step-by-step:
### Part (a)
Question: Which trigonometric ratio is represented by [tex]\( \frac{AB}{BC} \)[/tex] ?
Solution:
In a right-angled triangle, the tangent of an angle [tex]\(\theta\)[/tex] is defined as the ratio of the length of the opposite side to the length of the adjacent side. For the given triangle [tex]\(ABC\)[/tex], where [tex]\(\angle ACB = \theta\)[/tex] and [tex]\(\angle ABC = 90^\circ\)[/tex],
[tex]\[ \text{tan}(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AB}{BC} \][/tex]
Therefore, the trigonometric ratio represented by [tex]\( \frac{AB}{BC} \)[/tex] is [tex]\(\tan(\theta)\)[/tex].
### Part (b)
Question: What is the value of [tex]\( \left(\frac{AB}{AC}\right)^2 + \left(\frac{BC}{AC}\right)^2 \)[/tex]?
Solution:
This equation is a trigonometric identity derived from the Pythagorean theorem applied to the sides of a right-angled triangle.
[tex]\[ \left(\frac{AB}{AC}\right)^2 + \left(\frac{BC}{AC}\right)^2 = \sin^2(\theta) + \cos^2(\theta) \][/tex]
According to the Pythagorean identity in trigonometry:
[tex]\[ \sin^2(\theta) + \cos^2(\theta) = 1 \][/tex]
Therefore, the value of the given expression is:
[tex]\[ 1 \][/tex]
### Part (c)
Question: If [tex]\( AC = 20 \)[/tex] cm and [tex]\( \cos \theta = \frac{4}{5} \)[/tex], find the length of [tex]\( BC \)[/tex].
Solution:
Given:
[tex]\[ AC = 20 \, \text{cm} \][/tex]
[tex]\[ \cos \theta = \frac{4}{5} \][/tex]
First, we find [tex]\( AB \)[/tex]:
[tex]\[ \cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AB}{AC} \implies AB = AC \cdot \cos \theta = 20 \cdot \frac{4}{5} = 16 \, \text{cm} \][/tex]
Next, we use the Pythagorean theorem to find [tex]\( BC \)[/tex]:
[tex]\[ AC^2 = AB^2 + BC^2 \implies BC^2 = AC^2 - AB^2 \][/tex]
[tex]\[ BC^2 = 20^2 - 16^2 = 400 - 256 = 144 \implies BC = \sqrt{144} = 12 \, \text{cm} \][/tex]
Therefore, the length of [tex]\( BC \)[/tex] is:
[tex]\[ 12 \, \text{cm} \][/tex]
### Part (d)
Question: If the length of [tex]\( AC \)[/tex] were twice the length of [tex]\( AB \)[/tex], what would be the value of [tex]\( \theta \)[/tex] ?
Solution:
Let [tex]\(AC = 2 \cdot AB\)[/tex].
Given:
[tex]\[ AC = 2 \cdot AB \][/tex]
We can write:
[tex]\[ AB = \frac{AC}{2} \][/tex]
Now, let’s calculate [tex]\( \cos \theta \)[/tex]:
[tex]\[ \cos \theta = \frac{AB}{AC} = \frac{\frac{AC}{2}}{AC} = \frac{1}{2} \][/tex]
To find [tex]\( \theta \)[/tex], we use the arccosine function:
[tex]\[ \theta = \cos^{-1}\left(\frac{1}{2}\right) \][/tex]
In radians:
[tex]\[ \theta = \frac{\pi}{3} \, \text{radians} = 1.0471975511965979 \, \text{radians}\][/tex]
### Part (a)
Question: Which trigonometric ratio is represented by [tex]\( \frac{AB}{BC} \)[/tex] ?
Solution:
In a right-angled triangle, the tangent of an angle [tex]\(\theta\)[/tex] is defined as the ratio of the length of the opposite side to the length of the adjacent side. For the given triangle [tex]\(ABC\)[/tex], where [tex]\(\angle ACB = \theta\)[/tex] and [tex]\(\angle ABC = 90^\circ\)[/tex],
[tex]\[ \text{tan}(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AB}{BC} \][/tex]
Therefore, the trigonometric ratio represented by [tex]\( \frac{AB}{BC} \)[/tex] is [tex]\(\tan(\theta)\)[/tex].
### Part (b)
Question: What is the value of [tex]\( \left(\frac{AB}{AC}\right)^2 + \left(\frac{BC}{AC}\right)^2 \)[/tex]?
Solution:
This equation is a trigonometric identity derived from the Pythagorean theorem applied to the sides of a right-angled triangle.
[tex]\[ \left(\frac{AB}{AC}\right)^2 + \left(\frac{BC}{AC}\right)^2 = \sin^2(\theta) + \cos^2(\theta) \][/tex]
According to the Pythagorean identity in trigonometry:
[tex]\[ \sin^2(\theta) + \cos^2(\theta) = 1 \][/tex]
Therefore, the value of the given expression is:
[tex]\[ 1 \][/tex]
### Part (c)
Question: If [tex]\( AC = 20 \)[/tex] cm and [tex]\( \cos \theta = \frac{4}{5} \)[/tex], find the length of [tex]\( BC \)[/tex].
Solution:
Given:
[tex]\[ AC = 20 \, \text{cm} \][/tex]
[tex]\[ \cos \theta = \frac{4}{5} \][/tex]
First, we find [tex]\( AB \)[/tex]:
[tex]\[ \cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AB}{AC} \implies AB = AC \cdot \cos \theta = 20 \cdot \frac{4}{5} = 16 \, \text{cm} \][/tex]
Next, we use the Pythagorean theorem to find [tex]\( BC \)[/tex]:
[tex]\[ AC^2 = AB^2 + BC^2 \implies BC^2 = AC^2 - AB^2 \][/tex]
[tex]\[ BC^2 = 20^2 - 16^2 = 400 - 256 = 144 \implies BC = \sqrt{144} = 12 \, \text{cm} \][/tex]
Therefore, the length of [tex]\( BC \)[/tex] is:
[tex]\[ 12 \, \text{cm} \][/tex]
### Part (d)
Question: If the length of [tex]\( AC \)[/tex] were twice the length of [tex]\( AB \)[/tex], what would be the value of [tex]\( \theta \)[/tex] ?
Solution:
Let [tex]\(AC = 2 \cdot AB\)[/tex].
Given:
[tex]\[ AC = 2 \cdot AB \][/tex]
We can write:
[tex]\[ AB = \frac{AC}{2} \][/tex]
Now, let’s calculate [tex]\( \cos \theta \)[/tex]:
[tex]\[ \cos \theta = \frac{AB}{AC} = \frac{\frac{AC}{2}}{AC} = \frac{1}{2} \][/tex]
To find [tex]\( \theta \)[/tex], we use the arccosine function:
[tex]\[ \theta = \cos^{-1}\left(\frac{1}{2}\right) \][/tex]
In radians:
[tex]\[ \theta = \frac{\pi}{3} \, \text{radians} = 1.0471975511965979 \, \text{radians}\][/tex]
