An isosceles triangle has two sides of length w that make a 2á-degree angle. Write down two different formulas for the area of this triangle, in terms of w and á (Greek “alpha”). By equating the formulas, discover a relation involving sin 2á, siná, and cosá.
To find the area of an isosceles triangle with two sides of length w making a 2α-degree angle, we can use two different formulas:
Formula 1: Area = 0.5 * w^2 * sin(2α)
Formula 2: Area = 0.5 * w^2 * sin(α) * cos(α)
To discover a relation involving sin 2α, sin α, and cos α, we can equate these two formulas:
0.5 * w^2 * sin(2α) = 0.5 * w^2 * sin(α) * cos(α)
Now, let's simplify and solve for the relation involving sin 2α, sin α, and cos α:
Canceling out the common terms and dividing both sides by w^2:
sin(2α) = sin(α) * cos(α)
Using the double angle identity for sine:
2sin(α)cos(α) = sin(α) * cos(α)
Dividing both sides by sin(α) * cos(α) (assuming they are non-zero):
2 = 1
This relation shows that 2 is equal to 1, which is a contradiction. Therefore, there is a mistake in either deriving the formulas or in equating them.
To find two different formulas for the area of an isosceles triangle, we can use the fact that an isosceles triangle has two equal sides. Let's denote the length of these equal sides by 'w' and the angle they make by 'α'.
Formula 1:
The base of the triangle can be any one of the two equal sides. Let's choose the side with length 'w'. The height of the triangle can be found using trigonometry. The height divides the base into two equal parts, forming a right triangle. Using the sine function, we have:
sin α = (height) / w
Rearranging the equation, we find:
height = w * sin α
The area of the triangle can be calculated using the formula:
Area = 0.5 * base * height
Substituting the values, we get:
Area = 0.5 * w * w * sin α = 0.5 * w^2 * sin α
Formula 2:
In an isosceles triangle, the height is also the median and the perpendicular bisector of the base. We can use this fact to find another formula for the area.
The length of the median and the base form a right triangle with an angle of α/2. Using the cosine function, we have:
cos (α/2) = (median) / w
Rearranging the equation, we find:
median = w * cos (α/2)
The area of the triangle can also be calculated using the formula:
Area = 0.5 * base * median
Substituting the values, we get:
Area = 0.5 * w * w * cos (α/2) = 0.5 * w^2 * cos (α/2)
Now, let's equate the two formulas for the area:
0.5 * w^2 * sin α = 0.5 * w^2 * cos (α/2)
Canceling out the common factors, we have:
sin α = cos (α/2)
Using the double-angle identity, we know that sin 2α = 2 * sin α * cos α:
sin 2α = 2 * sin α * cos α
Replacing sin α with cos (α/2), we get:
sin 2α = 2 * cos (α/2) * cos α
This is the relation involving sin 2α, sin α, and cos α derived from equating the area formulas of the isosceles triangle.
For the whole triangle
Area = (1/2)(w)(w)sin 2a = (1/2)w^2sin 2a
draw in an altitude h, bisecting the angle 2a because we have an isosceles triangle
Area of triangle = (1/2)hwsin a + (1/2)whsina
= whsin a
but cos a = h/w
h = wcos a
then whsin a = w(wcos a)sin a = w^2sin a cos a
finally then
(1/2)w^2 sin 2a = w^2 sin a cos a
sin 2a = 2sin a cos a
nice!