Question

given an isosceles triangle with non each angle as 106° and non equal sides as 80 metres find the other side of the triangle

Answer 1

Isosceles Triangle

Formula: Let, $A$, $B$, $C$ be the vertices of the triangle $\Delta ABC$, and $a$, $b$, $c$ be the opposite sides respectively.

Then we have a formula:

$\quad\quad \frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}$

Solution:

• Given, the non-equal angle of the isosceles triangle is $106^{\circ}$.

• Let us take, $\angle A=106^{\circ}$

• Also given that the non-equal side is $80$ m.

• Then this is the opposite side of the vertex $A$.

• Let, $a=80$ m.

• It is clear that, the equal angles measure to:

• $\quad \frac{180^{\circ}-106^{\circ}}{2}=37^{\circ}$

• Then, $\angle B=\angle C=37^{\circ}$

• We have to find $b$ or $c$.

• Using the above formula, we get:

• $\frac{80}{sin106^{\circ}}=\frac{b}{sin37^{\circ}}=\frac{c}{sin37^{\circ}}$

• Since $b=c$, we take only the first two ratios:

• $\quad\frac{80}{sin106^{circ}}=\frac{b}{sin37^{\circ}}$

• $\Rightarrow b=\frac{80\:sin37^{\circ}}{sin106^{\circ}}$

• $\Rightarrow b=50.085$

• $\Rightarrow \boxed{b\approx 50.09}$

Answer: Therefore the length of the equal sides are $\bold{50.09}$ m each.