Question

If in a triangle ABC, \[b = \sqrt 3 \], c = 1, and \[B - C = {90^ \circ }\], then find \[\angle A\].
A. \[{30^ \circ }\]
B. \[{45^ \circ }\]
C. \[{75^ \circ }\]
D. \[{15^ \circ }\]

Answer 1

Hint: To solve the given question we will derive a formula \[\tan \dfrac{{B - C}}{2} = \dfrac{{b - c}}{{b + c}}\cot \dfrac{A}{2}\] by sine law for an oblique triangle, sum and difference of two sine of angles formula. Then we simply substitute the value of b, c, and B – C in the formula to get the required solution.



Formula Used:Sine Law:
\[\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}}\]
Sum of sine angles:
\[\sin A + \sin B = 2\sin \dfrac{{A + B}}{2}\cos \dfrac{{A - B}}{2}\]
Difference of sine angles:
\[\sin A - \sin B = 2\cos \dfrac{{A + B}}{2}\sin \dfrac{{A - B}}{2}\]



Complete step by step solution:We know the sine law
\[\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}}\]
Taking last two ratios of the sine law:
\[\dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}}\]
\[\dfrac{b}{c} = \dfrac{{\sin B}}{{\sin C}}\]
Applying componendo and dividendo rule:
\[ \Rightarrow \dfrac{{b + c}}{{b - c}} = \dfrac{{\sin B + \sin C}}{{\sin B - \sin C}}\]
Now applying Difference of sine angles and Sum of sine angles formula:
\[ \Rightarrow \dfrac{{b + c}}{{b - c}} = \dfrac{{2\sin \dfrac{{B + C}}{2}\cos \dfrac{{B - C}}{2}}}{{2\cos \dfrac{{B + C}}{2}\sin \dfrac{{B - C}}{2}}}\]
Cancel out 2 from denominator and numerator:
\[ \Rightarrow \dfrac{{b + c}}{{b - c}} = \dfrac{{\sin \dfrac{{B + C}}{2}\cos \dfrac{{B - C}}{2}}}{{\cos \dfrac{{B + C}}{2}\sin \dfrac{{B - C}}{2}}}\]
\[ \Rightarrow \dfrac{{b + c}}{{b - c}} = \tan \dfrac{{B + C}}{2}\cot \dfrac{{B - C}}{2}\]
We know that, \[B + C = {180^ \circ } - A\]
\[ \Rightarrow \dfrac{{b + c}}{{b - c}} = \tan \dfrac{{{{180}^ \circ } - A}}{2}\cot \dfrac{{B - C}}{2}\]
\[ \Rightarrow \dfrac{{b + c}}{{b - c}} = \tan \left( {{{90}^ \circ } - \dfrac{A}{2}} \right)\cot \dfrac{{B - C}}{2}\]
Apply formula of complementary angle formula \[\tan \left( {{{90}^ \circ } - \theta } \right) = \cot \theta \]
\[ \Rightarrow \dfrac{{b + c}}{{b - c}} = \cot \dfrac{A}{2}\cot \dfrac{{B - C}}{2}\]
\[ \Rightarrow \tan \dfrac{{B - C}}{2} = \dfrac{{b - c}}{{b + c}}\cot \dfrac{A}{2}\]
Now putting \[b = \sqrt 3 \], c = 1, and \[B - C = {90^ \circ }\]
\[ \Rightarrow \tan \dfrac{{{{90}^ \circ }}}{2} = \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}\cot \dfrac{A}{2}\]
\[ \Rightarrow \tan {45^ \circ } = \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}\cot \dfrac{A}{2}\]
\[ \Rightarrow 1 \cdot \tan \dfrac{A}{2} = \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}\]
\[ \Rightarrow \tan \dfrac{A}{2} = \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}\]
Rationalize the denominator:
 \[ \Rightarrow \tan \dfrac{A}{2} = \dfrac{{\left( {\sqrt 3 - 1} \right)\left( {\sqrt 3 - 1} \right)}}{{\left( {\sqrt 3 + 1} \right)\left( {\sqrt 3 - 1} \right)}}\]
\[ \Rightarrow \tan \dfrac{A}{2} = \dfrac{{3 + 1 - 2\sqrt 3 }}{{3 - 1}}\]
\[ \Rightarrow \tan \dfrac{A}{2} = 2 - \sqrt 3 \]
We know that \[\tan {15^ \circ } = 2 - \sqrt 3 \]
\[ \Rightarrow \tan \dfrac{A}{2} = \tan {15^ \circ }\]
\[ \Rightarrow \dfrac{A}{2} = {15^ \circ }\]
\[ \Rightarrow A = {30^ \circ }\]



Option ‘A’ is correct



Note: Students often do mistake the value of \[\tan {15^ \circ }\] and \[\tan {75^ \circ }\]. The value of \[\tan {15^ \circ }\] is \[2 - \sqrt 3 \]. The value of \[\tan {75^ \circ }\] is \[2 + \sqrt 3 \].