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.