ASA Formula for Area

Geometry Level 3

The formula for finding the area of a triangle based on just one side and the angles is:

A = a 2 sin ⁡ ( β + γ ) 4 ( 1 − cos ⁡ α ) A=\dfrac{a^2\sin(\beta+\gamma)}{4(1-\cos\alpha)} A = a 2 sin ⁡ α 8 cos ⁡ β cos ⁡ γ A=\dfrac{a^2\sin \alpha}{8\cos \beta\cos\gamma} A = sin ⁡ β sin ⁡ γ sin ⁡ α ( a 2 ) 2 A=\dfrac{ \sin \beta\sin\gamma}{\sin\alpha}\left(\dfrac a2\right)^2 A = a 2 sin ⁡ β sin ⁡ γ 2 sin ⁡ α A=\dfrac{a^2\sin \beta\sin\gamma}{2\sin\alpha} A = a 2 sin ⁡ β sin ⁡ γ 2 cos ⁡ α A=\dfrac{a^2\sin \beta\sin\gamma}{2\cos \alpha}

Marta Reece by Marta Reece , 69, USA

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1 solution

Marta Reece
May 30, 2017

Area of triangle based on two sides and their enclosed angle is: A = 1 2 a b sin ⁡ γ A=\dfrac 12 ab\sin\gamma

Side b b can be expressed in terms of a a using the law of sines: b sin ⁡ β = a sin ⁡ α \dfrac{b}{\sin\beta}=\dfrac{a}{\sin\alpha}

b = a sin ⁡ β sin ⁡ α b=\dfrac{a\sin\beta}{\sin\alpha}

Substituting that into the equation for area results is a formula

A = 1 2 a a sin ⁡ β sin ⁡ α sin ⁡ γ = a 2 sin ⁡ β sin ⁡ γ sin ⁡ α A=\dfrac 12 a\dfrac{a\sin\beta}{\sin\alpha}\sin\gamma=\dfrac {a^2\sin\beta\sin\gamma }{\sin\alpha}

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