Question

If the length of the sides of a triangle be 7, \[4\sqrt 3 \], and \[\sqrt {13} \], then find the smallest angle.
A. \[{15^ \circ }\]
B. \[{30^ \circ }\]
C. \[{60^ \circ }\]
D. \[{45^ \circ }\]

Answer 1

Hint First we apply cosine law to find the angles of the triangle. Then compare the angles to find the smallest angle.

Formula used:
Cosine law:
\[{a^2} = {b^2} + {c^2} - 2bc\cos A\]
\[{b^2} = {a^2} + {c^2} - 2ac\cos B\]
\[{c^2} = {a^2} + {b^2} - 2ab\cos C\]

Complete step by step solution
Given that the length of the sides of a triangle are 7, \[4\sqrt 3 \], and \[\sqrt {13} \].
Assume that, a = 7, b = \[4\sqrt 3 \], and c =\[\sqrt {13} \].
Now calculate angle A using the formula \[{a^2} = {b^2} + {c^2} - 2bc\cos A\]:
Substitute a = 7, b = \[4\sqrt 3 \], and c =\[\sqrt {13} \]:
\[{7^2} = {\left( {4\sqrt 3 } \right)^2} + {\left( {\sqrt {13} } \right)^2} - 2 \cdot \left( {4\sqrt 3 } \right) \cdot \left( {\sqrt {13} } \right) \cdot \cos A\]
\[ \Rightarrow 49 = 48 + 13 - 8\sqrt {39} \cos A\]
\[ \Rightarrow 48 + 13 - 49 = 8\sqrt {39} \cos A\]
\[ \Rightarrow 12 = 8\sqrt {39} \cos A\]
Divide both sides by \[8\sqrt {39} \]
\[ \Rightarrow \dfrac{{12}}{{8\sqrt {39} }} = \cos A\]
\[ \Rightarrow \dfrac{3}{{2\sqrt {39} }} = \cos A\]
\[ \Rightarrow \dfrac{{\sqrt 3 }}{{2\sqrt {13} }} = \cos A\]
\[ \Rightarrow A \approx {76.102^ \circ }\]
Now calculate angle B using the formula \[{b^2} = {a^2} + {c^2} - 2ac\cos B\]:
Substitute a = 7, b = \[4\sqrt 3 \], and c =\[\sqrt {13} \]:
\[{\left( {4\sqrt 3 } \right)^2} = {7^2} + {\left( {\sqrt {13} } \right)^2} - 2 \cdot \left( 7 \right) \cdot \left( {\sqrt {13} } \right) \cdot \cos B\]
\[ \Rightarrow 48 = 49 + 13 - 14\sqrt {13} \cos B\]
\[ \Rightarrow 49 + 13 - 48 = 14\sqrt {13} \cos B\]
\[ \Rightarrow 14 = 14\sqrt {13} \cos B\]
Divide both sides by \[14\sqrt {13} \]
\[ \Rightarrow \dfrac{{14}}{{14\sqrt {13} }} = \cos B\]
Cancel 14 from both sides:
\[ \Rightarrow \dfrac{1}{{\sqrt {13} }} = \cos B\]
\[ \Rightarrow B \approx {73.898^ \circ }\]
Now calculate angle B using the formula \[{c^2} = {a^2} + {b^2} - 2ab\cos C\]:
Substitute a = 7, b = \[4\sqrt 3 \], and c =\[\sqrt {13} \]:
\[{\left( {\sqrt {13} } \right)^2} = {\left( {4\sqrt 3 } \right)^2} + {7^2} - 2 \cdot 4\sqrt 3 \cdot 7 \cdot \cos C\]
\[ \Rightarrow 13 = 48 + 49 - 56\sqrt 3 \cos C\]
\[ \Rightarrow 48 + 49 - 13 = 56\sqrt 3 \cos C\]
\[ \Rightarrow 84 = 56\sqrt 3 \cos C\]
Divide both sides by \[56\sqrt 3 \]
\[ \Rightarrow \dfrac{{84}}{{56\sqrt 3 }} = \cos C\]
Divide denominator and numerator by 28
\[ \Rightarrow \dfrac{3}{{2\sqrt 3 }} = \cos C\]
\[ \Rightarrow \dfrac{{\sqrt 3 }}{2} = \cos C\]
\[ \Rightarrow C = {30^ \circ }\]
The angles of the triangle are \[{76.102^ \circ },{73.898^ \circ },{30^ \circ }\].
The smallest angle of the triangle is \[{30^ \circ }\].
Hence option B is the correct.

Note A triangle has three angles. So, we need to find all three angles of the triangle. Students often make mistakes to solve this question. They did not compute all angles of the triangle. But we need to find all three angles to find the smallest angle using this method. Either we can compute values of cosines of all three angles and then we can conclude that the largest cosine value will have smallest angle and then we can compute the value of angle.