I am stuck with the following problem :
What is the area of an isosceles triangle whose equal sides are $20$ cm and the angle between them is $30^{\circ}$ ?
It is a nineth standard problem and I can not use trigonometry(I mean I can not use the formula that involves sine ,cosine etc. rule) or integration.
One way to solve it as following :
Consider circumscribed circle and it's radius $R$. By inscribed angle theorem we can have , that $|c|=|R|$, where $c$ is third side of the triangle $a=b=10$. Now using formula $\displaystyle S=\frac{abc}{4R}$, where $S$ is area of triange. So:
$$S=\frac{20 \cdot 20 \cdot c}{4R}=\frac{400}{4}=100$$
Is there any other simpler way to tackle the problem? I also don't know how to use the angle as given in the question.
I will be highly obliged if someone gives a detailed clarification to the problem. Thanks in advance for your time.
$\endgroup$ 23 Answers
$\begingroup$Use:
In triangle $ABD:$
$$\angle D=90^{\circ},\angle A=30^{\circ} \Rightarrow BD=\frac12AB=10$$
$\endgroup$ 2 $\begingroup$Let $D$ be a point on $AC$ such that $AC\perp BD$, and let $B'\not =B$ be a point on $BD$ such that $DB=DB'$.
$\qquad\qquad\qquad$
Since $\triangle{ABB'}$ is an equilateral triangle, $BD=\frac 12BB'=\frac 12AB=10$.
$\endgroup$ 1 $\begingroup$How I would attack this: The isosceles triangle has two sides of length 20 cm and angle 30 degrees. You can use the cosine law to find the length of the base. A line from the vertex perpendicular to the base divides the triangle into two right triangles with one side half the base. The altitude of the isosceles triangle is given by the Pythagorean Theorem and then the area of the isosceles triangle is "1/2 base times height".
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