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Suppose you have an isosceles triangle, and each of the equal sides has a length of 1 foot. suppose the angle formed by those two sides is 45^\circ. then the area of the triangle is

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mathstudent55
The triangle has a 45-deg angle.
The base angles are congruent and measure 67.5 deg.
The congruent sides measure 1 ft.
Use law of sines to find the length of the base.

[tex] \dfrac{\sin 67.5^\circ}{1~ft} = \dfrac{\sin 45^\circ}{b} [/tex]

[tex] b = \dfrac{\sin 45^\circ}{\sin 67.5^\circ}~ft [/tex]

[tex] b = 0.765~ft [/tex]

Draw a height from the vertex of the 45-deg angle to the base.
Half of the base is 0.765 ft/2 = 0.383 ft

We can find the height of the triangle using the small triangles.

0.383^2 + h^2 = 1^2

h = 0.9239 ft

A = bh/2 = 0.765 * 0.9239/2 ft^2

A = 0.354 ft^2