Question
Find sinL, cosL, tanL, sinM, cosM, and tanM when ℓ=12, m=123⎯⎯√, and n=24. Match each ratio to the corresponding trigonometric expression.
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cosL=
tanL=
sinM=
cosM=
tanM=
sinL=
cosL=
tanL=
sinM=
cosM=
tanM=
Answers
GPT-4o mini
To find the values of sinL, cosL, tanL, sinM, cosM, and tanM given the right triangle with sides \( \ell = 12 \), \( m = 123\sqrt{m} \), and \( n = 24 \), we first need to clarify the lengths and their roles in the triangle.
Assuming the right triangle has:
- Opposite side to angle \( L \) as \( \ell = 12 \)
- Adjacent side to angle \( L \) as \( n = 24 \)
- Hypotenuse can be found using the Pythagorean theorem.
Let's denote the triangle sides as:
- \( a = \ell = 12 \) (opposite to angle \( L \))
- \( b = n = 24 \) (adjacent to angle \( L \))
- \( c = m = 123\sqrt{m} \) (hypotenuse)
1. First, we find the hypotenuse \( c \):
\[
c = \sqrt{a^2 + b^2} = \sqrt{12^2 + 24^2} = \sqrt{144 + 576} = \sqrt{720} = 12\sqrt{5}
\]
2. Now to find the trigonometric functions for angle \( L \):
- **Sine of angle \( L \)**:
\[
\sin L = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\ell}{c} = \frac{12}{12\sqrt{5}} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}
\]
- **Cosine of angle \( L \)**:
\[
\cos L = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{n}{c} = \frac{24}{12\sqrt{5}} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}
\]
- **Tangent of angle \( L \)**:
\[
\tan L = \frac{\text{opposite}}{\text{adjacent}} = \frac{\ell}{n} = \frac{12}{24} = \frac{1}{2}
\]
3. Now let’s find the trigonometric functions for angle \( M \):
For angle \( M \), we know:
- Opposite side to angle \( M \) is \( n = 24 \),
- Adjacent side to angle \( M \) is \( \ell = 12 \).
- **Sine of angle \( M \)**:
\[
\sin M = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{n}{c} = \frac{24}{12\sqrt{5}} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}
\]
- **Cosine of angle \( M \)**:
\[
\cos M = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\ell}{c} = \frac{12}{12\sqrt{5}} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}
\]
- **Tangent of angle \( M \)**:
\[
\tan M = \frac{\text{opposite}}{\text{adjacent}} = \frac{n}{\ell} = \frac{24}{12} = 2
\]
In summary:
- \(\sin L = \frac{\sqrt{5}}{5}\)
- \(\cos L = \frac{2\sqrt{5}}{5}\)
- \(\tan L = \frac{1}{2}\)
- \(\sin M = \frac{2\sqrt{5}}{5}\)
- \(\cos M = \frac{\sqrt{5}}{5}\)
- \(\tan M = 2\)
Assuming the right triangle has:
- Opposite side to angle \( L \) as \( \ell = 12 \)
- Adjacent side to angle \( L \) as \( n = 24 \)
- Hypotenuse can be found using the Pythagorean theorem.
Let's denote the triangle sides as:
- \( a = \ell = 12 \) (opposite to angle \( L \))
- \( b = n = 24 \) (adjacent to angle \( L \))
- \( c = m = 123\sqrt{m} \) (hypotenuse)
1. First, we find the hypotenuse \( c \):
\[
c = \sqrt{a^2 + b^2} = \sqrt{12^2 + 24^2} = \sqrt{144 + 576} = \sqrt{720} = 12\sqrt{5}
\]
2. Now to find the trigonometric functions for angle \( L \):
- **Sine of angle \( L \)**:
\[
\sin L = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\ell}{c} = \frac{12}{12\sqrt{5}} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}
\]
- **Cosine of angle \( L \)**:
\[
\cos L = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{n}{c} = \frac{24}{12\sqrt{5}} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}
\]
- **Tangent of angle \( L \)**:
\[
\tan L = \frac{\text{opposite}}{\text{adjacent}} = \frac{\ell}{n} = \frac{12}{24} = \frac{1}{2}
\]
3. Now let’s find the trigonometric functions for angle \( M \):
For angle \( M \), we know:
- Opposite side to angle \( M \) is \( n = 24 \),
- Adjacent side to angle \( M \) is \( \ell = 12 \).
- **Sine of angle \( M \)**:
\[
\sin M = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{n}{c} = \frac{24}{12\sqrt{5}} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}
\]
- **Cosine of angle \( M \)**:
\[
\cos M = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\ell}{c} = \frac{12}{12\sqrt{5}} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}
\]
- **Tangent of angle \( M \)**:
\[
\tan M = \frac{\text{opposite}}{\text{adjacent}} = \frac{n}{\ell} = \frac{24}{12} = 2
\]
In summary:
- \(\sin L = \frac{\sqrt{5}}{5}\)
- \(\cos L = \frac{2\sqrt{5}}{5}\)
- \(\tan L = \frac{1}{2}\)
- \(\sin M = \frac{2\sqrt{5}}{5}\)
- \(\cos M = \frac{\sqrt{5}}{5}\)
- \(\tan M = 2\)