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Trigonometric Functions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsTrigonometric Functions2 Marks
MCQ
In a triangle ABC , if $\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}$ then angle $C$ is equal to
  • A
    $30^{\circ}$
  • ✓
    $60^{\circ}$
  • C
    $90^{\circ}$
  • D
    $120^{\circ}$
✓

Answer

Correct option: B.
$60^{\circ}$
(B) $\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}$
$\Rightarrow \frac{ a + b +2 c }{( a + c )( b + c )}=\frac{3}{ a + b + c }$
$\begin{array}{l}\Rightarrow( a + b +2 c )( a + b + c )=3( a + c )( b + c ) \\ \Rightarrow a ^2+ b ^2- c ^2= ab \end{array}$
$\therefore \quad \cos C=\frac{a^2+b^2-c^2}{2 a b}=\frac{a b}{2 a b}=\frac{1}{2}=\cos 60^{\circ}$
$\Rightarrow C =60^{\circ}$

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