
In \[\Delta ABC\], if a = 16, b = 24, and c = 20, then find \[\cos \dfrac{B}{2}\].
A. \[\dfrac{3}{4}\]
B. \[\dfrac{1}{4}\]
C. \[\dfrac{1}{2}\]
D. \[\dfrac{1}{3}\]
Answer
162k+ views
Hint: First we will find the semi perimeter of the \[\Delta ABC\]. Then we apply the half angle cosine rule to calculate the value of \[\cos \dfrac{B}{2}\].
Formula used:
The semi perimeter of a triangle is \[s = \dfrac{{a + b + c}}{2}\], where a, b, c are length of sides.
The half angle cosine formula is \[\cos \dfrac{B}{2} = \sqrt {\dfrac{{s\left( {s - b} \right)}}{{ca}}} \]
Complete step by step solution:
Given that if a = 16, b = 24, and c = 20.
Now we will plug the value of a, b, c in \[s = \dfrac{{a + b + c}}{2}\] to calculate the semi perimeter.
\[s = \dfrac{{16 + 24 + 20}}{2}\]
\[ \Rightarrow s = \dfrac{{60}}{2}\]
\[ \Rightarrow s = 30\]
The half angle formula for cosine is \[\cos \dfrac{B}{2} = \sqrt {\dfrac{{s\left( {s - b} \right)}}{{ca}}} \].
Substitute the value of s, a, b, c:
\[\cos \dfrac{B}{2} = \sqrt {\dfrac{{30\left( {30 - 24} \right)}}{{20 \cdot 16}}} \]
\[ \Rightarrow \cos \dfrac{B}{2} = \sqrt {\dfrac{{30 \cdot 6}}{{20 \cdot 16}}} \]
\[ \Rightarrow \cos \dfrac{B}{2} = \sqrt {\dfrac{{3 \cdot 3}}{{16}}} \]
\[ \Rightarrow \cos \dfrac{B}{2} = \dfrac{3}{4}\]
Hence option A is the correct option.
Additional information:
We can derive the half angle formula by using trigonometry identities. We know cosine law:
\[{a^2} = {b^2} + {c^2} - 2bc\cos A\]
\[ \Rightarrow \cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\]
Now applying the trigonometry identity \[2{\cos ^2}\dfrac{A}{2} - 1 = \cos A\]
\[ \Rightarrow 2{\cos ^2}\dfrac{A}{2} - 1 = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\]
\[ \Rightarrow 2{\cos ^2}\dfrac{A}{2} = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}} + 1\]
\[ \Rightarrow 2{\cos ^2}\dfrac{A}{2} = \dfrac{{{b^2} + {c^2} - {a^2} + 2bc}}{{2bc}}\]
\[ \Rightarrow 2{\cos ^2}\dfrac{A}{2} = \dfrac{{{{\left( {b + c} \right)}^2} - {a^2}}}{{2bc}}\]
\[ \Rightarrow 2{\cos ^2}\dfrac{A}{2} = \dfrac{{\left( {b + c + a} \right)\left( {b + c - a} \right)}}{{2bc}}\]
\[ \Rightarrow 2{\cos ^2}\dfrac{A}{2} = \dfrac{{\left( {b + c + a} \right)\left( {b + c + a - 2a} \right)}}{{2bc}}\]
We know that \[2s = a + b + c\]
\[ \Rightarrow 2{\cos ^2}\dfrac{A}{2} = \dfrac{{2s\left( {2s - 2a} \right)}}{{2bc}}\]
\[ \Rightarrow {\cos ^2}\dfrac{A}{2} = \dfrac{{s\left( {s - a} \right)}}{{bc}}\]
Note: Students are often confused with semi perimeter and perimeter. The half of the perimeter is known as semi perimeter and semi perimeter is denoted by s. Perimeter is the sum of all sides.
Formula used:
The semi perimeter of a triangle is \[s = \dfrac{{a + b + c}}{2}\], where a, b, c are length of sides.
The half angle cosine formula is \[\cos \dfrac{B}{2} = \sqrt {\dfrac{{s\left( {s - b} \right)}}{{ca}}} \]
Complete step by step solution:
Given that if a = 16, b = 24, and c = 20.
Now we will plug the value of a, b, c in \[s = \dfrac{{a + b + c}}{2}\] to calculate the semi perimeter.
\[s = \dfrac{{16 + 24 + 20}}{2}\]
\[ \Rightarrow s = \dfrac{{60}}{2}\]
\[ \Rightarrow s = 30\]
The half angle formula for cosine is \[\cos \dfrac{B}{2} = \sqrt {\dfrac{{s\left( {s - b} \right)}}{{ca}}} \].
Substitute the value of s, a, b, c:
\[\cos \dfrac{B}{2} = \sqrt {\dfrac{{30\left( {30 - 24} \right)}}{{20 \cdot 16}}} \]
\[ \Rightarrow \cos \dfrac{B}{2} = \sqrt {\dfrac{{30 \cdot 6}}{{20 \cdot 16}}} \]
\[ \Rightarrow \cos \dfrac{B}{2} = \sqrt {\dfrac{{3 \cdot 3}}{{16}}} \]
\[ \Rightarrow \cos \dfrac{B}{2} = \dfrac{3}{4}\]
Hence option A is the correct option.
Additional information:
We can derive the half angle formula by using trigonometry identities. We know cosine law:
\[{a^2} = {b^2} + {c^2} - 2bc\cos A\]
\[ \Rightarrow \cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\]
Now applying the trigonometry identity \[2{\cos ^2}\dfrac{A}{2} - 1 = \cos A\]
\[ \Rightarrow 2{\cos ^2}\dfrac{A}{2} - 1 = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\]
\[ \Rightarrow 2{\cos ^2}\dfrac{A}{2} = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}} + 1\]
\[ \Rightarrow 2{\cos ^2}\dfrac{A}{2} = \dfrac{{{b^2} + {c^2} - {a^2} + 2bc}}{{2bc}}\]
\[ \Rightarrow 2{\cos ^2}\dfrac{A}{2} = \dfrac{{{{\left( {b + c} \right)}^2} - {a^2}}}{{2bc}}\]
\[ \Rightarrow 2{\cos ^2}\dfrac{A}{2} = \dfrac{{\left( {b + c + a} \right)\left( {b + c - a} \right)}}{{2bc}}\]
\[ \Rightarrow 2{\cos ^2}\dfrac{A}{2} = \dfrac{{\left( {b + c + a} \right)\left( {b + c + a - 2a} \right)}}{{2bc}}\]
We know that \[2s = a + b + c\]
\[ \Rightarrow 2{\cos ^2}\dfrac{A}{2} = \dfrac{{2s\left( {2s - 2a} \right)}}{{2bc}}\]
\[ \Rightarrow {\cos ^2}\dfrac{A}{2} = \dfrac{{s\left( {s - a} \right)}}{{bc}}\]
Note: Students are often confused with semi perimeter and perimeter. The half of the perimeter is known as semi perimeter and semi perimeter is denoted by s. Perimeter is the sum of all sides.
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