With the usual notation, in [ ]ABC, if [ A + B] = 120o, a = [ 3 ] + 1, b = [ 3 ] - 1 then …

Question

With the usual notation, in $$\Delta $$ABC, if  $$\angle A + \angle B$$ = 120o,  a = $$\sqrt 3 $$ $+$ 1, b = $$\sqrt 3 $$ - 1 then the ratio  $$\angle A:\angle B,$$ is :

Accepted Answer

A + B = 120o

$$\tan {{A - B} \over 2} = {{a - b} \over {a + b}}\cot \left( {{c \over 2}} \right)$$

$$ = {{\sqrt 3  + 1 - \sqrt 3  + 1} \over {2\left( {\sqrt 3 } \right)}}\cot \left( {{{30}^o}} \right) = {1 \over {\sqrt 3 }}.\sqrt 3  = 1$$

$${{A - B} \over 2} = {45^o}$$

$$ \Rightarrow A - B = {90^o}$$

$$ \ A + B = {120^o}$$

$2A = {210^o}$

$A = {105^o}$

$B = {15^o}$

$$ \therefore $$ $$\angle A:\angle B,$$ = 7 : 1

Solutions · 1