Triangle JKL is dilated by a scale factor of 2, which means that all the sides of the triangle will be doubled in length. Since we know the values of sin(b°) and cos(b°), we can use these trigonometric ratios to find tan(b°).
Let's start by finding the value of the hypotenuse in the original triangle JKL. We can use the Pythagorean theorem, which states that the square of the hypotenuse is equal to the sum of the squares of the other two sides.
In triangle JKL, we have:
sin(b°) = 3/5
cos(b°) = 4/5
Using the Pythagorean theorem:
hypotenuse^2 = (3/5)^2 + (4/5)^2
hypotenuse^2 = 9/25 + 16/25
hypotenuse^2 = 25/25
hypotenuse = 1
So, in the original triangle JKL, the hypotenuse has a length of 1.
Now, let's find the length of the hypotenuse in the dilated triangle. Since the triangle is dilated by a scale factor of 2, all the sides will be doubled. Therefore, the hypotenuse of the dilated triangle will have a length of 2.
Finally, we can find tan(b°) in the dilated triangle by using the formula tan(b°) = sin(b°) / cos(b°).
In the dilated triangle, we have:
sin(b°) = 3/5
cos(b°) = 4/5
tan(b°) = (3/5) / (4/5)
tan(b°) = 3/4
Therefore, in the dilated triangle JKL, tan(b°) is equal to 3/4.
To provide in summary after explanation:
- In the original triangle JKL, the hypotenuse has a length of 1.
- In the dilated triangle JKL, the hypotenuse has a length of 2.
- The tangent of angle b° in the dilated triangle JKL is 3/4.