Question

In a triangle ABC, AD is altitude from A. Given b>c,

$\angle C = 23 ^ { \circ }$ and $A D = \frac { a b c } { b ^ { 2 } - c ^ { 2 } }$ then $\angle B$ equal to

• A. $67 ^ { \circ }$
• B.
• C. $113 ^ { \circ }$
• D. None of these

Answer 1

Answer: (C)

$113 ^ { \circ }$

Answer 2

$\cos B = \frac { a ^ { 2 } + c ^ { 2 } - b ^ { 2 } } { 2 a c } = \frac { a ^ { 2 } - \left( b ^ { 2 } - c ^ { 2 } \right) } { 2 a c }$

Now, $A D = \frac { a b c } { b ^ { 2 } - c ^ { 2 } }$ ; $\therefore$ $\cos B = \frac { a ^ { 2 } - \frac { a b c } { A D } } { 2 a c }$

Also $A D = b \sin 23 ^ { \circ }$ ; $\therefore \cos B = \frac { a - \frac { c } { \sin 23 ^ { \circ } } } { 2 c }$

By sine formulae ⇒ $\frac { a } { c } = \frac { \sin \left( B + 23 ^ { \circ } \right) } { \sin 23 ^ { \circ } }$ ;

$\therefore$ $\cos B = \frac { \left[ \frac { \sin \left( B + 23 ^ { \circ } \right) } { \sin 23 ^ { \circ } } - \frac { 1 } { \sin 23 ^ { \circ } } \right] } { 2 }$

⇒ $\sin \left( 23 ^ { \circ } - B \right) = - 1 = \sin \left( - 90 ^ { \circ } \right)$ ; therefore $23 ^ { \circ } - B = - 90 ^ { \circ }$ or

$B = 113 ^ { \circ }$.