Question

Show that the area of an isosceles triangle is A=a^2 sin theta cos theta where a is the length of one of the two sides and theta is the measure of one of two equal angles.

Answer 1

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Solution

Answer 2

PROVED

Step-by-step explanation:

given that : ABC is an isosceles triangle

to prove :  $A = a^2 \sin\theta \cos\theta$

proof : we know that the 2 sides are equal in the isosceles triangle

therefore ,

AB = BC

thus , $\angle B = \angle C= \theta$

(angle opposite to equal sides are equal )

now , in triangle ABC

sum of all the sides is 180 degree

therefore ,

$\angle A + \angle B + \angle C = 180^\circ \\\\\angle A+ \theta +\theta = 180^\circ\\\\\angle A = 180 - 2\theta$

now , area of isosceles triangle

$\frac{1}{2}\times AB \times AC \times \sin \angle A \\\\\frac{1}{2}\times a \times a \times \sin (180-2\theta)\\\\\frac{1}{2}\times a^2 \times \sin 2(90-\theta)\\\\\frac{1}{2}\times a^2 \times 2 \sin (90-\theta)\cos(90-\theta )\\\\\frac{1}{2}\times a^2 \times 2\cos\theta\sin\theta \\\\= a^2\cos\theta \sin\theta\\\\A = a^2\cos\theta \sin\theta$

hence proved

#Learn more :

One of the angels of a triangle measures 80° and the other two angles are equal. find the measure of each of the equal angles.

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