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

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Answer 1 · 2016-06-22T16:47:30

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

Let us name the isosceles $DeltaABC$ , with legs $AB=AC=4.4cm$ , base $BC.$

Now, Area of $DeltaABC=1/2*AB*AC*sinA.$ Hence, we get the eqn.,

$1/2*AB*AC*sinA=4.$

$:. 1/2(4.4)(4.4)sinA=4.$

$:. (2.2)(4.4)sinA=4.$

$:. sinA=4/{(2.2)(4.4)}=1/{(2.2)(1.1)}=1/2.42~=0.4132$

$:. A=arc(sin0.4132)=24^0,24',$ using Table of Natural Sine.