Construct a right triangle ABC such that BC =4 cm and hypotenuse makes an angle of 30 degree at B. verify that AB =2 AC
so, did you do the construction? It is well known that for a 30-60-90 triangle, the hypotenuse is twice the shortest side.
To construct a right triangle ABC such that BC = 4 cm and the hypotenuse makes an angle of 30 degrees at B, you can follow these steps:
1. Draw a line segment BC of length 4 cm. This will serve as the base of the triangle.
- Label the starting point as B and the endpoint as C.
2. From point B, draw a line at an angle of 30 degrees to BC.
- Use a protractor to measure the angle or estimate it visually.
3. Mark the intersection point of the line and the segment BC as point A.
- Label the point of intersection as A.
4. Draw the line segments AB and AC.
- Join point A to points B and C with line segments.
Now, we need to verify that AB = 2AC.
To do this, we can use the concept of trigonometric ratios.
In the right triangle ABC:
- The hypotenuse is BC, which is equal to 4 cm.
- The side opposite the 30-degree angle is AC.
- The side adjacent to the 30-degree angle is AB.
We know that the sine function relates the ratio of the length of the opposite side to the length of the hypotenuse.
sin(angle) = opposite/hypotenuse
In our case, the sine of 30 degrees will be:
sin(30°) = AC/BC
sin(30°) = AC/4
We can solve this equation to find the length of AC:
AC = 4 * sin(30°)
AC = 4 * 0.5
AC = 2 cm
Now, to verify that AB = 2AC, we can substitute the value of AC that we found:
AB = 2 * AC
AB = 2 * 2
AB = 4 cm
So, AB is indeed equal to 2AC, which is 4 cm in this case.
To construct the right triangle ABC with BC = 4 cm and the hypotenuse making a 30-degree angle at B, follow these steps:
1. Draw a line segment BC of length 4 cm, which will be the base of the triangle.
2. From point B, construct a 30-degree angle. To do this, use a protractor to measure a 30-degree angle from the line segment BC. Mark the point where the angle intersects with BC as point D.
3. Draw the perpendicular bisector of BD. To do this, draw a line from point D to the midpoint of BC. Mark the point of intersection between the perpendicular bisector and BD as point E.
4. With point E as the center, construct a circle with a radius of BE. This circle will intersect with BC at point F.
5. Draw the line segment AF, which will be the hypotenuse of the right triangle.
6. Finally, draw the line segment AC, which will connect points A and C.
Now, to verify that AB = 2AC, we can use the properties of the right triangle. Let's label the length of AB as x and the length of AC as y.
Since AF is the hypotenuse and BC is the base, according to the Pythagorean theorem, we have:
AF^2 = AB^2 + BC^2
(AF)^2 = (x)^2 + (4)^2
(AF)^2 = x^2 + 16
Since the hypotenuse AF makes a 30-degree angle, we can use trigonometric ratios to find the values of AB and AC. The side opposite the 30-degree angle is AB, and the side adjacent to the 30-degree angle is AC. Therefore, we have:
tan(30°) = AB/AC
1/√3 = x/y [tan(30°) = 1/√3]
Now, let's solve the two equations simultaneously:
(AF)^2 = x^2 + 16
1/√3 = x/y
Substituting the second equation into the first equation:
(AF)^2 = (y)^2 + 16/3 [Squaring both sides & replacing x/y with 1/√3]
(AF)^2 = (y)^2 +16/3
(AF)^2 - (y)^2 = 16/3
Now, substituting AF with x+y (since AF is the hypotenuse and equals AB + AC) and simplifying:
(x+y)^2 - (y)^2 = 16/3
x^2 + 2xy + y^2 - y^2 = 16/3
x^2 + 2xy = 16/3
Substituting the value of y from the second equation in the above equation:
x^2 + 2x(√3x) = 16/3 or x^2 + 2√3x^2 = 16/3
(1 + 2√3)x^2 = 16/3
x^2 = (16/3) / (1 + 2√3)
Now we can solve for x:
x^2 = (16/3) / (1 + 2√3)
x^2 = (16/3) * (1 - 2√3) / (1 - 2√3)
x^2 = 16(1 - 2√3) / (3)(1 - 2√3)
x^2 = 16 - 32√3 + 48 / (3 - 6√3)
x^2 = 64 / (3 - 6√3)
Simplifying further:
x^2 = 64(3 + 6√3) / (9 - 108)
x^2 = 64(1/2√3 + √3) / (1 - 12)
x^2 = 64(1/2√3 + √3) / (-11)
x^2 = -64 * (1/2√3 + √3) / 11
x^2 = -64(1/2√3 + √3) / 11
x^2 = -32√3 - 64 / 11
Since we cannot square root a negative number as the side of a triangle, we conclude that there is no real value for x. Therefore, AB is not equal to 2AC.