hello.
how to find the base of an isosceles triangle if the equal sides are given and the angle between them too.
for example... Q- find the base of the isosceles triangle if the equal sides are 6 cm each and the angle between them is 45 degree.
( try not to involve trigonometry )
Anonymous:
You may use Pythagoras theorem
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With out use of sine or cosine of angles, it seems not possible to find it easily.
Let ABC be the isosceles triangle, with A being the intersection of equal sides. So angle A = 45°. angles B = C = (180°-45°)/2 = 67.5°.
Given AB = AC = 6 cm.
The easy way is :
AB / Sin C = BC / Sin A = 2 R = circumcircle diameter
=> BC = AB * Sin A / Sin C = 2 R * Sin A
BC = base = 6 * Sin 45° / Sin 67.5° = 4.592 cm
=============================================
Let ABC be the isosceles triangle, with A being the intersection of equal sides. So angle A = 45°. angles B = C = (180°-45°)/2 = 67.5°.
Given AB = AC = 6 cm.
The easy way is :
AB / Sin C = BC / Sin A = 2 R = circumcircle diameter
=> BC = AB * Sin A / Sin C = 2 R * Sin A
BC = base = 6 * Sin 45° / Sin 67.5° = 4.592 cm
=============================================
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