Question

Show that the area of an Issosceles triangle is A = a Sin 0 Cos 0.
where a is the length of
one of the two equal sides and is the measure
of one of two equal angles​

Answer 1

Answer:

given that : ABC is an isosceles triangle

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

2

sinθcosθ

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

therefore ,

AB = BC

thus , \angle B = \angle C= \theta∠B=∠C=θ

(angle opposite to equal sides are equal )

now , in triangle ABC

sum of all the sides is 180 degree

therefore ,

\begin{gathered}\angle A + \angle B + \angle C = 180^\circ \\\\\angle A+ \theta +\theta = 180^\circ\\\\\angle A = 180 - 2\theta\end{gathered}

∠A+∠B+∠C=180

∘

∠A+θ+θ=180

∘

∠A=180−2θ

now , area of isosceles triangle

\begin{gathered}\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\end{gathered}

2

1

×AB×AC×sin∠A

2

1

×a×a×sin(180−2θ)

2

1

×a

2

×sin2(90−θ)

2

1

×a

2

×2sin(90−θ)cos(90−θ)

2

1

×a

2

×2cosθsinθ

=a

2

cosθsinθ

A=a

2

cosθsinθ