Welcome to Westonci.ca, the place where your questions find answers from a community of knowledgeable experts. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
Sagot :
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]
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.