What is the formula to find an angle when you have 3 sides of a triangle \[a=13m,b=8.5m,c=9m\]?
Answer
565.5k+ views
Hint: We know that if a, b and c are sides of a triangle then\[\cos A=\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}\], \[\operatorname{cosB}=\dfrac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}\] and \[\operatorname{cosC}=\dfrac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}\]. By using this concept, the given concept can be solved.
Complete step by step solution:
From the question, it is clear that we have to find an angle when we are given 3 sides of a triangle \[a=13m,b=8.5m,c=9m\].
We know that if a, b and c are sides of a triangle then\[\cos A=\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}\], \[\operatorname{cosB}=\dfrac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}\] and \[\operatorname{cosC}=\dfrac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}\].
So, by using the above formulae we can find the cosine of angle A, B and C of a triangle.
Let us assume
\[\begin{align}
& a=13m.....(1) \\
& b=8.5m.....(2) \\
& c=9m.....(3) \\
\end{align}\]
Let us assume
\[\cos A=\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}......(4)\]
Now let us substitute equation (1), equation (2) and equation (3) in equation (4), then we get
\[\begin{align}
& \Rightarrow \cos A=\dfrac{{{(8.5)}^{2}}+{{9}^{2}}-{{13}^{2}}}{2(8.5)(9)} \\
& \Rightarrow \cos A=\dfrac{\dfrac{289}{4}+81-169}{153} \\
& \Rightarrow \cos A=\dfrac{\dfrac{289}{4}-88}{153} \\
& \Rightarrow \operatorname{cosA}=\dfrac{-7}{68} \\
\end{align}\]
So, it is clear that \[\operatorname{cosA}=\dfrac{-7}{68}\].
Let us assume this as equation (5), then we get
\[\operatorname{cosA}=\dfrac{-7}{68}.......(5)\]
From equation (5), we get
\[\Rightarrow A={{\cos }^{-1}}\left( \dfrac{-7}{68} \right)\]
Let us assume this as equation this as equation (6).
\[A={{\cos }^{-1}}\left( \dfrac{-7}{68} \right)......(6)\]
We know that
\[{{\cos }^{-1}}(-x)=\pi -{{\cos }^{-1}}x\]
Now let us assume the above concept to equation (6), then we get
\[\Rightarrow A=\pi -{{\cos }^{-1}}\left( \dfrac{7}{68} \right)....(7)\]
From equation (7), it is clear that\[A=\pi -{{\cos }^{-1}}\left( \dfrac{7}{68} \right)\].
In this way we can find an angle when you have 3 sides of a triangle \[a=13m,b=8.5m,c=9m\].
Note: Students may have a misconception that if a, b and c are sides of a triangle then\[\cos A=\dfrac{{{b}^{2}}+{{c}^{2}}+{{a}^{2}}}{2bc}\] but we know that if a, b and c are sides of a triangle then\[\cos A=\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}\]. Students should avoid these misconceptions to have a final answer not to get interrupted. Students should also avoid calculation mistakes while solving this problem.
Complete step by step solution:
From the question, it is clear that we have to find an angle when we are given 3 sides of a triangle \[a=13m,b=8.5m,c=9m\].
We know that if a, b and c are sides of a triangle then\[\cos A=\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}\], \[\operatorname{cosB}=\dfrac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}\] and \[\operatorname{cosC}=\dfrac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}\].
So, by using the above formulae we can find the cosine of angle A, B and C of a triangle.
Let us assume
\[\begin{align}
& a=13m.....(1) \\
& b=8.5m.....(2) \\
& c=9m.....(3) \\
\end{align}\]
Let us assume
\[\cos A=\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}......(4)\]
Now let us substitute equation (1), equation (2) and equation (3) in equation (4), then we get
\[\begin{align}
& \Rightarrow \cos A=\dfrac{{{(8.5)}^{2}}+{{9}^{2}}-{{13}^{2}}}{2(8.5)(9)} \\
& \Rightarrow \cos A=\dfrac{\dfrac{289}{4}+81-169}{153} \\
& \Rightarrow \cos A=\dfrac{\dfrac{289}{4}-88}{153} \\
& \Rightarrow \operatorname{cosA}=\dfrac{-7}{68} \\
\end{align}\]
So, it is clear that \[\operatorname{cosA}=\dfrac{-7}{68}\].
Let us assume this as equation (5), then we get
\[\operatorname{cosA}=\dfrac{-7}{68}.......(5)\]
From equation (5), we get
\[\Rightarrow A={{\cos }^{-1}}\left( \dfrac{-7}{68} \right)\]
Let us assume this as equation this as equation (6).
\[A={{\cos }^{-1}}\left( \dfrac{-7}{68} \right)......(6)\]
We know that
\[{{\cos }^{-1}}(-x)=\pi -{{\cos }^{-1}}x\]
Now let us assume the above concept to equation (6), then we get
\[\Rightarrow A=\pi -{{\cos }^{-1}}\left( \dfrac{7}{68} \right)....(7)\]
From equation (7), it is clear that\[A=\pi -{{\cos }^{-1}}\left( \dfrac{7}{68} \right)\].
In this way we can find an angle when you have 3 sides of a triangle \[a=13m,b=8.5m,c=9m\].
Note: Students may have a misconception that if a, b and c are sides of a triangle then\[\cos A=\dfrac{{{b}^{2}}+{{c}^{2}}+{{a}^{2}}}{2bc}\] but we know that if a, b and c are sides of a triangle then\[\cos A=\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}\]. Students should avoid these misconceptions to have a final answer not to get interrupted. Students should also avoid calculation mistakes while solving this problem.
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