Question

In triangle $ABC,AB=AC=10cm.\angle ABC={50}^{\circ }$.
(a) Find the length of $BC$.
(b) Find the diameter of the circle.
$[\sin {50}^{\circ }=0.77,\cos {50}^{\circ }=0.64,\tan {50}^{\circ }=1.19]$

Answer 1

Consider the given figure:
Since, $AB=AC=10cm,△ABC$ is an isosceles triangle and hence the base angles are equal.
i.e. $m\angle B=m\angle C={50}^{\circ }$
By angle sum property, we have $m\angle A+m\angle B+m\angle C={180}^{\circ }$
i.e. $m\angle A+{50}^{\circ }+{50}^{\circ }={180}^{\circ }$
i.e. $m\angle A+{100}^{\circ }={180}^{\circ }\Rightarrow m\angle A={80}^{\circ }$
The perpendicular from the vertex of an isosceles triangle to the base bisects the base.
(a) Consider the right triangle $△ABD$:
$\cos {50}^{\circ }=\frac{BD}{AB}\Rightarrow 0.64=\frac{BD}{10}$
$\Rightarrow 0.64\times 10=BD$
$\Rightarrow BD=6.4cm$
We have, $BC=BD+DC=25BD=2\times 6.4=12.8cm$.
(b) From the figure, it is clear that $OA=OB=OC$ is the circumradius of $△ABC$.
Let $A$ be the area of $△ABC$ and let $R$ be the circumradius.
We know that $A=\frac{abc}{4R}$, where $a,b,c$ are the sides of the triangle.
From $△ABC$,
$\tan 50=\frac{AD}{BD}\Rightarrow 1.19=\frac{AD}{6.4}$
$\Rightarrow AD=1.19\times 6.4=7.616cm$
Area $\left(△ABC\right)=\frac{1}{2}\times base\times height=\frac{1}{2}\times BC\times AD=\frac{1}{2}\times 12.8\times 7.616=48.7424sq.cm$
Since, $A=\frac{abc}{4R}$, we have $R=\frac{abc}{4A}\Rightarrow R=\frac{10\times 10\times 12.8}{4\times 48.7424}=6.57cm$
Thus, diameter of the circle $=2R=2\times 6.57=13.14cm$