Question

With the usual notation, in $\Delta ABC$, if $\angle A + \angle B = 120^\circ ,a = \sqrt 3 + 1$ and $b = \sqrt 3 - 1$. Then, what will be the ratio for $\angle A:\angle B$ ?

Answer 1

This question is from Trigonometric geometry. We have to know the angle property for any triangle will be $\angle A + \angle B + \angle C = 180^\circ$. By using sine rule property for triangle ABC, we have to show $\dfrac{{a - b}}{{a + b}}\cot \dfrac{C}{2} = \tan \left( {\dfrac{{A - B}}{2}} \right)$. Then, by putting given values of a and b in this equation we will get $\angle A:\angle B$

Complete step-by-step answer:

Given, $\angle A + \angle B = 120^\circ {\rm{ }}...{\rm{(i)}}$

{\rm{In }}\Delta {\rm{ABC, by using Sine Rule,we get}}\\\ \dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}} = k \end{array}$$ Now, we can write equation for a, b and c $$a = k\sin A,b = k\sin C,c = k\sin C$$ Now, we have $$ \Rightarrow \dfrac{{a - b}}{{a + b}}\cot \dfrac{C}{2}$$ We can write values of a, b and c in this equation $$ = \left( {\dfrac{{k\sin A - k\sin B}}{{k\sin A + k\sin B}}} \right)\cot \dfrac{C}{2}$$ We can expand this equation as follows, $$ = \dfrac{{2\cos \left( {\dfrac{{A + B}}{2}} \right).\sin \left( {\dfrac{{A - B}}{2}} \right)}}{{2\sin \left( {\dfrac{{A + B}}{2}} \right).\cos \left( {\dfrac{{A - B}}{2}} \right)}}.\dfrac{{\cos \left( {\dfrac{C}{2}} \right)}}{{\sin \left( {\dfrac{C}{2}} \right)}}$$ But, we know $$\begin{array}{l} A + B + C = \pi \\\ A + B = \pi - C \end{array}$$ Now, we can put values of A + B in this equation $$ \Rightarrow \dfrac{{\cos \left( {\dfrac{\pi }{2} - \dfrac{C}{2}} \right).\sin \left( {\dfrac{{A - B}}{2}} \right)}}{{\sin \left( {\dfrac{\pi }{2} - \dfrac{C}{2}} \right).\cos \left( {\dfrac{{A - B}}{2}} \right)}} \times \dfrac{{\cos \left( {\dfrac{C}{2}} \right)}}{{\sin \left( {\dfrac{C}{2}} \right)}}$$ On simplification, we get $$ \Rightarrow \dfrac{{\sin \left( {\dfrac{C}{2}} \right)}}{{\sin \left( {\dfrac{C}{2}} \right)}} \times \tan \left( {\dfrac{{A - B}}{2}} \right) \times \dfrac{{\cos \left( {\dfrac{C}{2}} \right)}}{{\sin \left( {\dfrac{C}{2}} \right)}}$$ $$ \Rightarrow \tan \left( {\dfrac{{A - B}}{2}} \right)$$ Hence, we get $$ \Rightarrow \dfrac{{a - b}}{{a + b}}\cot \dfrac{C}{2} = \tan \left( {\dfrac{{A - B}}{2}} \right)$$ Now, $$\tan \left( {\dfrac{{A - B}}{2}} \right) = \dfrac{{a - b}}{{a + b}}\cot \left( {\dfrac{C}{2}} \right)$$ We can put given values of a = and b = in this equation. $$ = \dfrac{{\sqrt 3 + 1 - \sqrt 3 + 1}}{{2\left( {\sqrt 3 } \right)}}\cot \left( {\dfrac{{60^\circ }}{2}} \right)$$ $$\left[ \begin{array}{l} {\rm{As }}A + B + C = 180^\circ \\\ \angle C = 180^\circ - 120^\circ = 60^\circ \end{array} \right]$$ $$ = \dfrac{{\sqrt 3 + 1 - \sqrt 3 + 1}}{{2\left( {\sqrt 3 } \right)}}\cot \left( {30^\circ } \right)$$ On simplification, we get $$ = \dfrac{1}{{\sqrt 3 }}.\sqrt 3 = 1$$ $$ \Rightarrow \tan \left( {\dfrac{{A - B}}{2}} \right) = 1$$ Now, we can convert this equation as, $$\dfrac{{A - B}}{2} = {\tan ^{ - 1}}\left( 1 \right)$$ But, we know, $${\tan ^{ - 1}}\left( 1 \right) = 45^\circ $$ Now, equation becomes $$ \Rightarrow A - B = 90^\circ $$ We can write this equation in angle form, $$ \Rightarrow \angle A - \angle B = 90^\circ {\rm{ }}...{\rm{(ii)}}$$ By adding Eqs.(i) and (ii), we get $$\angle 2A = 210^\circ $$ On simplification, we get $$ \Rightarrow \angle A = \dfrac{{210^\circ }}{2} = 105^\circ $$ After putting the value of $$\angle A$$ in Eq.(i). We get $$\begin{array}{l} 105^\circ + \angle B = 120^\circ \\\ \Rightarrow \angle B = 120^\circ - 105^\circ = 15^\circ \end{array}$$ This is the required solution. **So, the correct answer is “Option A”.** **Note:** Calculation plays an important role in these types of trigonometric problems. Students should know the angle property that the sum of all internal angles of a triangle is always $$180^\circ $$. In this problem, while using formulas student can do mistake like $$\tan \left( {\dfrac{{A - B}}{2}} \right) = \dfrac{{a + b}}{{a - b}}\cot \left( {\dfrac{C}{2}} \right)$$ instead of $$\tan \left( {\dfrac{{A - B}}{2}} \right) = \dfrac{{a - b}}{{a + b}}\cot \left( {\dfrac{C}{2}} \right)$$ Here, students must take while using trigonometric formulae. There is an alternate method to solve this question. We have sine rule, $$\dfrac{{\sqrt 3 + 1}}{{\sin \left( {120 - x} \right)}} = \dfrac{{\sqrt 3 - 1}}{{\sin x}}$$ Now, we can rewrite this equation as, $$ \Rightarrow \dfrac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}} = \dfrac{{\sin \left( {120 - x} \right)}}{{\sin x}}$$ $$\begin{array}{l} \Rightarrow \tan x = 2 - \sqrt 3 \\\ \Rightarrow x = 15^\circ \\\ \therefore \dfrac{{\angle A}}{{\angle B}} = \dfrac{7}{1} \end{array}$$