MCQ
In triangles $\text{ABC}$ and $\text{DEF} , \angle\text{A}=\angle\text{E}=40^\circ, AB : ED = AC : EF$ and $\angle\text{F}=65^\circ,$ then $\angle\text{B}=$
- A$35^\circ$
- B$65^\circ$
- ✓$75^\circ$
- D$85^\circ$
✓
Answer
Correct option: C.
$75^\circ$
In $\triangle\text{ABC}$ and $\triangle\text{DEF},$
$\angle\text{A}=\angle\text{E}=40^\circ$
$AB : ED = AC : EF, \angle\text{F}=65^\circ$
$\Rightarrow\frac{\text{AB}}{\text{ED}}=\frac{\text{AC}}{\text{EF}}$
$\because$ In $\triangle\text{ABC}$ and $\triangle\text{EDF},$
$\angle\text{A}=\angle\text{E} \ ($each $= 40^\circ )$
$\frac{\text{AB}}{\text{ED}}=\frac{\text{AC}}{\text{EF}} \ ($given$)$
$\therefore\triangle\text{ABC}\sim\triangle\text{EDF} \ (\text{SAS}$ criterion$)$
$\therefore\angle\text{C}=\angle\text{F}=65^\circ$
and $\angle\text{B}=\angle\text{D}$
But $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ ($Sum of angles of a triangle$)$
$\Rightarrow40^\circ+65^\circ+\angle\text{C}=180^\circ$
$\Rightarrow105^\circ+\angle\text{C}=180^\circ$
$\therefore\angle\text{C}=180^\circ-105^\circ=75^\circ$
$\angle\text{A}=\angle\text{E}=40^\circ$
$AB : ED = AC : EF, \angle\text{F}=65^\circ$
$\Rightarrow\frac{\text{AB}}{\text{ED}}=\frac{\text{AC}}{\text{EF}}$
$\because$ In $\triangle\text{ABC}$ and $\triangle\text{EDF},$
$\angle\text{A}=\angle\text{E} \ ($each $= 40^\circ )$
$\frac{\text{AB}}{\text{ED}}=\frac{\text{AC}}{\text{EF}} \ ($given$)$
$\therefore\triangle\text{ABC}\sim\triangle\text{EDF} \ (\text{SAS}$ criterion$)$
$\therefore\angle\text{C}=\angle\text{F}=65^\circ$
and $\angle\text{B}=\angle\text{D}$
But $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ ($Sum of angles of a triangle$)$
$\Rightarrow40^\circ+65^\circ+\angle\text{C}=180^\circ$
$\Rightarrow105^\circ+\angle\text{C}=180^\circ$
$\therefore\angle\text{C}=180^\circ-105^\circ=75^\circ$
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