Question

In the given figure, $AD||BC,\angle AFG=b$ and $\angle BCD=c$. Express $b$ in terms of $c$.

• A. $\frac{c}{2}$
• B. $\frac{{90}^{\circ }+c}{2}$
• C. ${180}^{\circ }-\frac{c}{2}$
• D. ${90}^{\circ }-\frac{c}{2}$

Answer 1

The correct option is D ${90}^{\circ }-\frac{c}{2}$

In $△FED$,
$\angle DFE=\angle AFG=b$ (vert. opp. $\angle s$)
$\angle BEC=\angle DFE=b$ (base $\angle s$ of an isos. $\Delta$)
$\angle FDC=b+b=2b$ .....(i) (ext $\angle =$ sum of int. opp. $\angle s$)
Also, $\angle FDC={180}^{\circ }-\angle BCD={180}^{\circ }-c$ (ii) ($\because AD||BC$, co-int $\angle s$)
$\therefore$ From (i) and (ii), $2b={180}^{\circ }-c$
$\Rightarrow b=\frac{{180}^{\circ }-c}{2}={90}^{\circ }-\frac{c}{2}$