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

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}}$$.