Question

In the above figure, $DEF$ is a triangle whose side $DF$ is produced to $G.HF$ and $HD$ are the bisectors of $\angle EFG$ and $\angle EDG$, respectively. If $\angle DEF=23{\frac{1}{2}}^{\circ }$, then $\angle DHF$ (in degrees) $=$

• A. $11\frac{1}{2}$
• B. $11\frac{2}{5}$
• C. $11\frac{3}{4}$
• D. $11\frac{1}{3}$

Answer 1

Answer: (C)

(i) $DH$ and $FH$ are angle bisectors. Exterior angle of a triangle equals the sum of its interior opposite angles.
(ii) Take $\angle FDH=\angle HDE=x$ and $\angle EFH=\angle HFG$ $=\gamma$.
(iii) $\angle EFG$ is the exterior angle of $△DEF$.
(iv) $\angle DEF=23{\frac{1}{2}}^{\circ }$ and $2x+y=180^{\circ }-23{\frac{1}{2}}^{\circ }$
(v) Find $x$ and $y$ and $H$.
$\therefore 2y=2x+23\frac{1}{2}$
Simplify to obtain the value of the required angle.