Find the perimeter of (a) triangle ABE who's mesurement is the side are 5/2 CM and 3 3/5 CM and the base is 2 3/4cm
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there is. It's called the "Law of cosines" and can be aplied in each and every triangle.
Law of cosines: In the ABC triangle,
\displaystyle BC^{2}=AB^{2}+AC^{2}-2XABXACBC2=AB2+AC2−2XABXAC
Let's consider that our ABC triangle has the measures in the following way:
AB=3; BC=5; AC=4
Applying the Law of cosines:
\displaystyle 5^{2}=3^{2}+4^{2}-2X3X4XcosAcos A=0 => A=90^\circ => sinA=152=32+42−2X3X4XcosAcosA=0=>A=90∘=>sinA=1
An area formula is ABXACXsinA / 2.
So, the ABC's area is 3X4X1/2=6
Hope I helped you.
Law of cosines: In the ABC triangle,
\displaystyle BC^{2}=AB^{2}+AC^{2}-2XABXACBC2=AB2+AC2−2XABXAC
Let's consider that our ABC triangle has the measures in the following way:
AB=3; BC=5; AC=4
Applying the Law of cosines:
\displaystyle 5^{2}=3^{2}+4^{2}-2X3X4XcosAcos A=0 => A=90^\circ => sinA=152=32+42−2X3X4XcosAcosA=0=>A=90∘=>sinA=1
An area formula is ABXACXsinA / 2.
So, the ABC's area is 3X4X1/2=6
Hope I helped you.
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