Question

The sides of triangle are \[3x + 4y\], \[4{\rm{x}} + 3{\rm{y }}\,{\rm{and}}\;\,5x + 5{\rm{y}}\] units, where \[{\rm{x}},{\rm{y}} > 0\] . Name the type of the triangle.
A. Right angled
B. Obtuse angled
C. Equilateral
D. None of these

Answer 1

Hint: First we will check whether given triangle is an equilateral triangle to compare the sides of the triangle. Then we will find the largest angle by using cosine formula to check whether the given triangle obtuse triangle or right angled triangle.




Formula Used:Cosine formula:
\[\cos {\rm{C }} = \,\,\dfrac{{{{\rm{a}}^2} + {b^2} - {{\rm{c}}^2}}}{{2{\rm{ab}}}}\]



Complete step by step solution:The lengths of sides of the triangle are \[3x + 4y\], \[4{\rm{x}} + 3{\rm{y }}\,{\rm{and}}\;\,5x + 5{\rm{y}}\] units.
Assume that, \[a = 3x + 4y\], \[b = 4x + 3y\], and \[c = 5x + 5y\]


Image: Triangle ABC
Since the lengths of the sides of the triangle is not equal to each other. Thus the triangle is not an equilateral triangle.
If the value of x and y are equal, then the largest side of the triangle is c. Thus the largest angle of the triangle is angle C.
Now we will find the value of angle C using cosine rule:
\[\cos {\rm{C }} = \,\,\dfrac{{{{\rm{a}}^2} + {b^2} - \,{{\rm{c}}^2}}}{{2{\rm{ab}}}}\]

Putting the values of a, b, and c in the formula,


\[\cos {\rm{C }} = \,\,\dfrac{{{{\left( {3{\rm{x}} + 4y} \right)}^2} + {{\left( {4{\rm{x}} + 3{\rm{y}}} \right)}^2} - {{\left( {5{\rm{x}} + 5y} \right)}^2}}}{{2 \times \left( {3{\rm{x}} + 4y} \right) \times \left( {4{\rm{x}} + 3{\rm{y}}} \right)}}\]

Applying algebraically identity: \[\cos {\rm{C }} = \,\,\dfrac{{{{\left( {3{\rm{x}}} \right)}^2} + {{\left( {4y} \right)}^2} + 2 \times \left( {3{\rm{x}}} \right)\left( {4y} \right) + {{\left( {4{\rm{x}}} \right)}^2} + {{\left( {3{\rm{y}}} \right)}^2} + 2 \times \left( {4{\rm{x}}} \right)\left( {3y} \right) - \left[ {{{\left( {5{\rm{x}}} \right)}^2} + {{\left( {5{\rm{y}}} \right)}^2} + 2 \times \left( {5{\rm{x}}} \right)\left( {5y} \right)} \right]}}{{\left( {6{\rm{x}} + 8y} \right) \times \left( {4{\rm{x}} + 3{\rm{y}}} \right)}}\]Simplify the above expression:

\[\cos {\rm{C }} = \,\,\dfrac{{9{{\rm{x}}^2} + 16{{\rm{y}}^2} + 24{\rm{xy}} + 16{{\rm{x}}^2} + 9{y^2} + 24{\rm{xy}} - \left[ {25{{\rm{x}}^2} + 25{{\rm{y}}^2} + 50{\rm{xy}}} \right]}}{{6x \times \left( {4{\rm{x}} + 3{\rm{y}}} \right) + 8y\left( {4{\rm{x}} + 3{\rm{y}}} \right)}}\]


 \[\cos {\rm{C }} = \,\,\dfrac{{9{{\rm{x}}^2} + 16{{\rm{y}}^2} + 24{\rm{xy}} + 16{{\rm{x}}^2} + 9{y^2} + 24{\rm{xy}} - 25{{\rm{x}}^2} - 25{{\rm{y}}^2} - 50{\rm{xy}}}}{{6x \times \left( {4{\rm{x}} + 3{\rm{y}}} \right) + 8y\left( {4{\rm{x}} + 3{\rm{y}}} \right)}}\]


\[ = \dfrac{{ - \,2{\rm{xy}}}}{{2\left( {12{{\rm{x}}^2} + 25xy + 12{{\rm{y}}^2}} \right)}}\] \( < 0\)

Therefore, C is an obtuse angle.



Option ‘B’ is correct



Note: Another procedure to check it:
Check by putting \[{\rm{x}} = 1\] and \[{\rm{y}} = 1\]
\[\cos {\rm{C}} = \dfrac{{ - \,2 \times 1 \times 1}}{{2\left( {12 \times 1 \times 1 + 25 \times 1 \times 1 + 12 \times 1 \times 1} \right)}}\]

\(\cos {\rm{C}}\,\, = \,\,\left( {\dfrac{{ - 1}}{{49}}} \right)\)
\(\cos \,{\rm{C}}\,\, = \,\, - 0\,.020\)
\[{\rm{C}}\, = \,\,{\cos ^{ - 1}}\,\,\,\left( { - 0\,.020} \right)\,\,\]
\({\rm{C}}\,\, = \,271.169^\circ \)

The value of the inverse of cosine\(\, - 0\,.020\) is\(271.169^\circ \). It simply tells that the angle is obtuse.