Hello, cllynn213!
In a rhomboid the obtuse angle is \(\displaystyle 123^o\), the longer side is \(\displaystyle 50\) mm, and the longer diagonal is \(\displaystyle 80\) mm.
Find the shorter side of the rhomboid
Code:
B
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A 50 C
We are given: \(\displaystyle \,A\,=\,123^o,\:AC\,=\,50,\:BC\,=\,80\)
\(\displaystyle \;\;\)and we want: \(\displaystyle x\,=\,AB\).
Using the Law of Sines:
\(\displaystyle \L\;\;\;\frac{\sin B}{50}\,=\,\frac{\sin123^o}{80}\)\(\displaystyle \;\;\Rightarrow\;\;\sin B\,=\,0.524169105\;\;\Rightarrow\;\;B\,\approx\,31.6^o\)
Then: \(\displaystyle \,C\:=\:180^o\,-\,123^o\,-\,31.6^o\:=\:25.4^o\)
Hence: \(\displaystyle \L\,\frac{x}{\sin25.4^o}\,=\,\frac{80}{\sin123^o}\)\(\displaystyle \;\;\Rightarrow\;\;x\,\approx\,40.9\)
Using the Law of Cosines:
\(\displaystyle \;\;\;80^2\:=\:x^2\,+\,50^2\,-\,2\cdot x\cdot50\cdot\cos123^o\)
We have the quadratic: \(\displaystyle \,x^2\,+\,54.46x\,-\,3900\:=\:0\)
Quadratic Formula: \(\displaystyle \L\,x\:=\:\frac{-54.46\,\pm\,\sqrt{54.46^2\,+\,4(3900)}}{2(1)} \:= \:\frac{-54.46\,\pm\,136.2567}{2}\)
\(\displaystyle \;\;\;\)Therefore: \(\displaystyle \,x\,\approx\,40.9\)
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