Question
these have to be equal please find the answers that make m=them equal Express the trig ratios as fractions in simplest terms.
I
J
K
39
65
52
Answer
Attempt 1 out of 2
sin KK =
cos JJ =
sin KK and cos JJ
.
Submit Answer
I
J
K
39
65
52
Answer
Attempt 1 out of 2
sin KK =
cos JJ =
sin KK and cos JJ
.
Submit Answer
Answers
GPT-4o mini
To make the sine of angle \( K \) equal to the cosine of angle \( J \), we need to find the angles that will satisfy the relationship \( \sin(K) = \cos(J) \).
Using the co-function identity, we know that:
\[
\sin(K) = \cos(J) \implies K + J = 90^\circ
\]
Let’s denote the angles as follows:
- \( K = 39^\circ \)
- \( J = 51^\circ \) (since \( 90^\circ - 39^\circ = 51^\circ \))
Now let’s express those trig ratios as fractions:
1. **For \( \sin(K) \), where \( K = 39^\circ \)**:
Using the sine function, you may directly use a calculator or sine tables, but to express it as a fraction in simplest form depends on the specific triangles involved if we were to find one corresponding to this specific angle:
\[
\sin(39^\circ) \quad \text{(can be evaluated numerically)}
\]
But in terms of general theory:
If it were part of a special triangle, we'd express it as a fraction, but generally \( K \) does not relate directly to a simple fraction without additional geometry.
2. **For \( \cos(J) \), where \( J = 51^\circ \)**:
Again, using cosine directly:
\[
\cos(51^\circ) \quad \text{(similarly can be evaluated numerically)}
\]
However, based on the co-function identities as noted above, it must be that:
\[
\sin(39^\circ) = \cos(51^\circ)
\]
Upon further check, using the relationship \( K + J = 90^\circ \), we can confirm that \( K = 39^\circ \) and \( J = 51^\circ \) makes them equal.
### Summary:
- \( \sin(K) = \sin(39^\circ) \)
- \( \cos(J) = \cos(51^\circ) \)
And they are indeed equal since \( 39^\circ + 51^\circ = 90^\circ \) confirming \( \sin(39^\circ) = \cos(51^\circ) \).
If you wish to proceed solving specifically for those numerical fractions or simpler forms like decimals, please let me know!
Using the co-function identity, we know that:
\[
\sin(K) = \cos(J) \implies K + J = 90^\circ
\]
Let’s denote the angles as follows:
- \( K = 39^\circ \)
- \( J = 51^\circ \) (since \( 90^\circ - 39^\circ = 51^\circ \))
Now let’s express those trig ratios as fractions:
1. **For \( \sin(K) \), where \( K = 39^\circ \)**:
Using the sine function, you may directly use a calculator or sine tables, but to express it as a fraction in simplest form depends on the specific triangles involved if we were to find one corresponding to this specific angle:
\[
\sin(39^\circ) \quad \text{(can be evaluated numerically)}
\]
But in terms of general theory:
If it were part of a special triangle, we'd express it as a fraction, but generally \( K \) does not relate directly to a simple fraction without additional geometry.
2. **For \( \cos(J) \), where \( J = 51^\circ \)**:
Again, using cosine directly:
\[
\cos(51^\circ) \quad \text{(similarly can be evaluated numerically)}
\]
However, based on the co-function identities as noted above, it must be that:
\[
\sin(39^\circ) = \cos(51^\circ)
\]
Upon further check, using the relationship \( K + J = 90^\circ \), we can confirm that \( K = 39^\circ \) and \( J = 51^\circ \) makes them equal.
### Summary:
- \( \sin(K) = \sin(39^\circ) \)
- \( \cos(J) = \cos(51^\circ) \)
And they are indeed equal since \( 39^\circ + 51^\circ = 90^\circ \) confirming \( \sin(39^\circ) = \cos(51^\circ) \).
If you wish to proceed solving specifically for those numerical fractions or simpler forms like decimals, please let me know!