# How do you find an angle in an isosceles triangle with sides length 4.4cm and area 4cm^2?

Jun 22, 2016

Vertex Angle $= {24}^{0} , 24 ' .$

#### Explanation:

Let us name the isosceles $\Delta A B C$, with legs $A B = A C = 4.4 c m$, base $B C .$

Now, Area of $\Delta A B C = \frac{1}{2} \cdot A B \cdot A C \cdot \sin A .$ Hence, we get the eqn.,

$\frac{1}{2} \cdot A B \cdot A C \cdot \sin A = 4.$

$\therefore \frac{1}{2} \left(4.4\right) \left(4.4\right) \sin A = 4.$

$\therefore \left(2.2\right) \left(4.4\right) \sin A = 4.$

$\therefore \sin A = \frac{4}{\left(2.2\right) \left(4.4\right)} = \frac{1}{\left(2.2\right) \left(1.1\right)} = \frac{1}{2.42} \cong 0.4132$

$\therefore A = a r c \left(\sin 0.4132\right) = {24}^{0} , 24 ' ,$ using Table of Natural Sine.