Question
Question
A right triangle has a hypotenuse of length 40 and an angle of 25°, with a side opposite this angle of length 16. A second right triangle also has an angle of 25°, with a hypotenuse of length 10. Determine the length of the side opposite the 25° angle on the second triangle.(1 point)
Responses
The side opposite the second triangle’s 25° angle has length 4.
The side opposite the second triangle’s 25° angle has length 4.
The side opposite the second triangle’s 25° angle has length 64.
The side opposite the second triangle’s 25° angle has length 64.
The side opposite the second triangle’s 25° angle has length 8.
The side opposite the second triangle’s 25° angle has length 8.
The side opposite the second triangle’s 25°angle has length 821−−√
.
The side opposite the second triangle’s 25°angle has length 8 Start Root 21 End Root .
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A right triangle has a hypotenuse of length 40 and an angle of 25°, with a side opposite this angle of length 16. A second right triangle also has an angle of 25°, with a hypotenuse of length 10. Determine the length of the side opposite the 25° angle on the second triangle.(1 point)
Responses
The side opposite the second triangle’s 25° angle has length 4.
The side opposite the second triangle’s 25° angle has length 4.
The side opposite the second triangle’s 25° angle has length 64.
The side opposite the second triangle’s 25° angle has length 64.
The side opposite the second triangle’s 25° angle has length 8.
The side opposite the second triangle’s 25° angle has length 8.
The side opposite the second triangle’s 25°angle has length 821−−√
.
The side opposite the second triangle’s 25°angle has length 8 Start Root 21 End Root .
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Answers
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Answered by
GPT-4o mini
To find the length of the side opposite the 25° angle in the second triangle, we can use the properties of right triangles and trigonometric ratios.
In any right triangle, the sine of an angle is given by the ratio of the length of the side opposite the angle to the length of the hypotenuse. For the right triangle, we know:
\[
\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}
\]
For the first triangle, where the hypotenuse is 40 and the opposite side (to the 25° angle) is 16:
\[
\sin(25^\circ) = \frac{16}{40}
\]
Now we'll simplify the right side:
\[
\sin(25^\circ) = \frac{16}{40} = \frac{2}{5} = 0.4
\]
Using the sine ratio in the second triangle where the hypotenuse is 10:
\[
\sin(25^\circ) = \frac{x}{10}
\]
Where \( x \) is the length of the side opposite the 25° angle in the second triangle. We can substitute the value of \(\sin(25^\circ)\) we found from the first triangle:
\[
0.4 = \frac{x}{10}
\]
Now we can solve for \( x \):
\[
x = 10 \times 0.4 = 4
\]
Thus, the length of the side opposite the second triangle’s 25° angle is:
\[
\boxed{4}
\]
In any right triangle, the sine of an angle is given by the ratio of the length of the side opposite the angle to the length of the hypotenuse. For the right triangle, we know:
\[
\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}
\]
For the first triangle, where the hypotenuse is 40 and the opposite side (to the 25° angle) is 16:
\[
\sin(25^\circ) = \frac{16}{40}
\]
Now we'll simplify the right side:
\[
\sin(25^\circ) = \frac{16}{40} = \frac{2}{5} = 0.4
\]
Using the sine ratio in the second triangle where the hypotenuse is 10:
\[
\sin(25^\circ) = \frac{x}{10}
\]
Where \( x \) is the length of the side opposite the 25° angle in the second triangle. We can substitute the value of \(\sin(25^\circ)\) we found from the first triangle:
\[
0.4 = \frac{x}{10}
\]
Now we can solve for \( x \):
\[
x = 10 \times 0.4 = 4
\]
Thus, the length of the side opposite the second triangle’s 25° angle is:
\[
\boxed{4}
\]
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