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the length of the hypotenuse of a 30-60-90 triangle is 4. what is the perimeter? the answer choices are 4+12√3, 6+2√3, 2+6√3, and 12 +4√3.
In a 30-60-90 triangle, the sides are in the ratio 1 : √3 : 2. Let x be the length of the shorter leg (opposite the 30 degree angle), then:
- The longer leg (opposite the 60 degree angle) has length √3x.
- The hypotenuse has length 2x (given as 4 in this case).
Therefore, we can set up the equation 2x = 4, and solve for x:
- 2x = 4
- x = 2
Using this value, we can find the lengths of the other two sides:
- The longer leg has length √3x = √3(2) = 2√3.
- The perimeter is the sum of all three sides: 2 + 2√3 + 4 = 6 + 2√3.
Therefore, the answer is option B: 6 + 2√3.
what are the values of w and x in the triangle below? round the answers to the nearest tenth. triangle ABC, ab is 14. bc is w, and the angle ACB is 38. triangle ABD's angle of ADB is 19.
To solve for the values of w and x in the triangle below, we can use the Law of sines.
First, we can find the measure of angle ABC by using the property that the angles in a triangle sum up to 180 degrees.
angle ABC = 180 - angle ACB - angle ADB
angle ABC = 180 - 38 - 19
angle ABC = 123
Then, we can use the Law of sines to solve for w:
(w / sin(38)) = (14 / sin(123))
w = 14(sin(38)) / sin(123)
w ≈ 7.5
Next, we can use the Law of sines again to solve for x:
(x / sin(19)) = (14 / sin(123))
x = 14(sin(19)) / sin(123)
x ≈ 4.7
Therefore, the values of w and x are approximately 7.5 and 4.7, respectively.
I got w= 10.9, and x = 15.8, how is this possible?
It's possible that there may have been a mistake in the calculation process. To confirm the correct answer, we can use the Law of sines again:
w / sin(angle ABC) = 14 / sin(angle ACB)
x / sin(angle ABD) = 14 / sin(angle ACB)
We already found that angle ABC ≈ 123 and angle ABD ≈ 19. Plugging these in and solving for w and x, we get:
w ≈ 9.6
x ≈ 4.2
Therefore, the correct values of w and x, rounded to one decimal place, are approximately 9.6 and 4.2, respectively.
what is the image of O(-3,-2) after two reflections, first across the line y=-5, then across the line x=1?
To find the image of point O(-3,-2) after two reflections, first across the line y=-5 and then across the line x=1, we can use the following steps:
Step 1: Reflect point O across the line y=-5.
The line y=-5 is the horizontal line passing through y=-5 on the y-axis.
The distance between O and the line y=-5 is 3 units. After reflection across the line, the distance between the image point O' and the line y=-5 should also be 3 units. Therefore, the y-coordinate of O' is -5-3=-8.
Thus, the image of point O after reflection across the line y=-5 is O'(-3,-8).
Step 2: Reflect point O' across the line x=1.
The line x=1 is the vertical line passing through x=1 on the x-axis.
The distance between O' and the line x=1 is 4 units. After reflection across the line, the distance between the final image point O'' and the line x=1 should also be 4 units. Therefore, the x-coordinate of O'' is 1-4=-3, and the final y-coordinate is -8.
Thus, the final image of point O after two reflections is O''(-3,-8).
Therefore, the image of point O(-3,-2) after two reflections, first across the line y=-5, then across the line x=1 is O''(-3,-8).