Question

In a triangle \[\vartriangle ABC\] if \[a\cos A = b\cos B\], then prove that the triangle is either a right angled triangle or an isosceles triangle.

Answer 1

Hint: Here we write the values \[a,b\] in terms of angles opposite to them using the Law of sines which states that Ratio of length of a side of a triangle to the sine of the angle opposite to that side is same for all sides and angles of a triangle and then solve using the trigonometric formulas.
* Sum of all three angles of a triangle is always \[{180^ \circ }\].
* Right angled triangle is a triangle where one angle is \[{90^ \circ }\].
* An isosceles triangle is a triangle having two sides equal to each other, also the angles opposite to equal sides are equal.

Complete step by step solution:
We draw a \[\vartriangle ABC\] with angles \[A,B,C\] and sides \[a,b,c\].

Since, we know sum of three angles of a triangle is \[{180^ \circ }\]
\[A + B + C = {180^ \circ }\]
Therefore, equation can be written as
 \[
  A + B = {180^ \circ } - C \\
  B + C = {180^ \circ } - A \\
  C + A = {180^ \circ } - B \\
 \] \[...(i)\]
From the diagram, angle opposite to the side \[a\] is \[A\], angle opposite to the side \[b\] is \[B\], angle opposite to the side \[c\] is \[C\].
Therefore, from Law of Sines.
\[\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c} = k\](say)
Taking LCM on right side we can write
\[a = \dfrac{{\sin A}}{k},b = \dfrac{{\sin B}}{k},c = \dfrac{{\sin C}}{k}\] \[...(ii)\]
 Substitute values from equation \[(ii)\] in \[a\cos A = b\cos B\] and solve.
\[\dfrac{{\sin A}}{k} \times \cos A = \dfrac{{\sin B}}{k} \times \cos B\]
Cancel out the denominator on both sides as it is equal on both sides.
\[\sin A\cos A = \sin B\cos B\]
Multiply both sides of the equations by \[2\]
\[2\sin A\cos A = 2\sin B\cos B\]
Using the trigonometric identity \[\sin 2\theta = 2\sin \theta \cos \theta \] we can write,
\[\sin 2A = \sin 2B\]
\[\sin 2A - \sin 2B = 0\]
Using the trigonometric identity \[\sin \alpha - \sin \beta = 2\cos \left( {\dfrac{{\alpha + \beta }}{2}} \right).\sin \left( {\dfrac{{\alpha - \beta }}{2}} \right)\]
Therefore \[\sin 2A - \sin 2B = 2\cos \left( {\dfrac{{2A + 2B}}{2}} \right).\sin \left( {\dfrac{{2A - 2B}}{2}} \right) = 0\]
\[2\cos \left( {\dfrac{{2(A + B)}}{2}} \right).\sin \left( {\dfrac{{2(A - B)}}{2}} \right) = 0\]
\[2\cos \left( {A + B} \right).\sin \left( {A - B} \right) = 0\]
Substitute the value of \[A + B = {180^ \circ } - C\] from equation \[(i)\]
\[2\cos \left( {{{180}^ \circ } - C} \right).\sin \left( {A - B} \right) = 0\]
Now we know \[\cos ({180^ \circ } - C) = - \cos C\]because cosine is negative in the fourth quadrant.

\[2( - \cos C).\sin \left( {A - B} \right) = 0\]
\[\cos C.\sin \left( {A - B} \right) = 0\]
If product of two numbers is zero implies either one of them is zero or both of them is zero.
Taking \[\cos C = 0\]
We know \[\cos \dfrac{\pi }{2} = 0\]
Therefore, \[C = \dfrac{\pi }{2}\]
Which means angle \[C\] is \[{90^ \circ }\]. Therefore, \[\vartriangle ABC\] is a right angled triangle.
Now, taking \[\sin (A - B) = 0\]
We know \[\sin {0^ \circ } = 0\]
Therefore, \[A - B = 0\]
\[A = B\]
Therefore, in \[\vartriangle ABC\], two angles are equal. Therefore, \[\vartriangle ABC\] is an isosceles triangle.
Note:
Students are likely to make mistakes while calculating the values like \[\cos ({180^ \circ } - C)\], they can refer to the quadrant diagram for easy conversions. Also, ratio in law of sines holds both ways, therefore \[\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c}\] is same as \[\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}}\].