the base of a trapezoid are 22 and 12 respectively. the angle at the extremities of one base are 65 degree and 45 degree respectively. find two legs?
label the parallelogram ABCD, where AD = 12, BC = 22
angle B = 45° and angle C = 65°
draw a line DE || AB to meet BC at E
Thus, BE = 12, and EC = 10
label AB = DE = x
and DC = y
clearly, angle DEC = 45°
In triangle DEC,
sin45/y = sin65/x = sin70/10
sin45/y = sin70/10
ysin70 = 10sin45
y = 10sin45/sin70 = appr 7.525
sin65/x = sin70/10
xsin70 = 10sin65
x = 10sin65/sin70 = appr 9.645
To find the lengths of the legs of a trapezoid given the bases and angles, you can use the Law of Cosines.
Let's call the longer base of the trapezoid "b1" and the shorter base "b2". In this case, b1 = 22 and b2 = 12.
Let's also call the angle at the end of b1 "θ1" and the angle at the end of b2 "θ2". In this case, θ1 = 65 degrees and θ2 = 45 degrees.
Now, we can use the Law of Cosines formula, which states:
c^2 = a^2 + b^2 - 2ab * cos(C)
Where c is the side we want to solve for, a and b are the adjacent sides, and C is the angle opposite side c.
In our case, we want to find the lengths of the legs, so we can label the adjacent sides to the angles as a1, a2, and the legs as c1 and c2.
For c1:
a1 = b1 and b1 = 22
b1 = b2 and b2 = 12
C = 65 degrees
So, using the Law of Cosines:
c1^2 = a1^2 + b1^2 - 2 * a1 * b1 * cos(C)
c1^2 = 22^2 + 12^2 - 2 * 22 * 12 * cos(65)
c1^2 = 484 + 144 - 528 * cos(65)
c1^2 ≈ 452.62
c1 ≈ √452.62
c1 ≈ 21.27
Similarly, for c2:
a2 = b1 and b1 = 22
b2 = b2 and b2 = 12
C = 45 degrees
So, using the Law of Cosines:
c2^2 = a2^2 + b2^2 - 2 * a2 * b2 * cos(C)
c2^2 = 22^2 + 12^2 - 2 * 22 * 12 * cos(45)
c2^2 = 484 + 144 - 528 * cos(45)
c2^2 ≈ 292.63
c2 ≈ √292.63
c2 ≈ 17.09
Therefore, the lengths of the legs of the trapezoid are approximately 21.27 and 17.09.
To find the two legs of a trapezoid, we can use the properties of the trapezoid and some trigonometry.
Let's label the shorter base of the trapezoid as 'a' and the longer base as 'b'. The angles at the extremities of the shorter base are given as 65 degrees and 45 degrees.
First, let's find the height of the trapezoid. The height of a trapezoid is the perpendicular distance between the two bases. The height can be found using the formula:
height = (b - a) * tan(angle)
In this case, the shorter base 'a' is 12, the longer base 'b' is 22, and the angle is 65 degrees. Plugging in these values, we have:
height = (22 - 12) * tan(65)
Using a calculator, we find that the height is approximately 8.323.
Next, let's find the lengths of the two legs of the trapezoid. The legs are the non-parallel sides of the trapezoid. We can find the lengths of the legs using the Pythagorean theorem.
The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides.
In this case, we have a right triangle with one leg as the height of the trapezoid (8.323) and the other leg as the difference between the longer base (22) and the shorter base (12):
length of leg = sqrt((b - a)^2 + height^2)
For the leg opposite the angle of 65 degrees:
length of leg 1 = sqrt((22 - 12)^2 + 8.323^2)
Using a calculator, we find that the length of the first leg is approximately 17.169.
For the leg opposite the angle of 45 degrees:
length of leg 2 = sqrt((12 - 22)^2 + 8.323^2)
Again, using a calculator, we find that the length of the second leg is also approximately 17.169.
Therefore, the lengths of the two legs of the trapezoid are approximately 17.169 each.