A) \[{{67}^{o}}\]
B) \[{{44}^{o}}\]
C) \[{{113}^{o}}\]
D) None of these
Correct Answer: C
Solution :
\[\cos B=\frac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}=\frac{{{a}^{2}}-({{b}^{2}}-{{c}^{2}})}{2ac}\] Now, \[AD=\frac{abc}{{{b}^{2}}-{{c}^{2}}}\]; \ \[\cos B=\frac{{{a}^{2}}-\frac{abc}{AD}}{2ac}\] Also, \[AD=b\sin {{23}^{o}}\]; \ \[\cos B=\frac{a-\frac{c}{\sin {{23}^{o}}}}{2c}\] By sine formula, \[\frac{a}{c}=\frac{\sin (B+{{23}^{o}})}{\sin {{23}^{o}}}\] \\[\cos B=\left( \frac{\sin (B+{{23}^{o}})}{\sin {{23}^{o}}}-\frac{1}{\sin {{23}^{o}}} \right)\div 2\] Þ \[\sin ({{23}^{o}}-B)=-1=\sin (-{{90}^{o}})\] \ \[{{23}^{o}}-B=-{{90}^{o}}\]or \[B={{113}^{o}}\].
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