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MODEL PAPER 1 (BASIC) — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsMODEL PAPER 1 (BASIC)1 Mark
MCQ
In triangles ABC and DEF, $\angle A =\angle E =40^{\circ}, AB : ED = AC : EF$ and $\angle F =65^{\circ}$, then $\angle B =$
  • A
    $75^{\circ}$
  • B
    $85^{\circ}$
  • C
    $35^{\circ}$
  • D
    $65^{\circ}$
✓

Answer

(A) $75^{\circ}$
Explanation: In $\triangle ABC$ and $\triangle DEF$
$\angle A=\angle E=40^{\circ}$
$A B: E D=A C: E F, \angle F=65^{\circ}$
Image
$\Rightarrow \frac{ AB }{ ED }=\frac{ AC }{ EF }$
$\because$ In $\triangle ABC$ and $\triangle EDF$
$\angle A =\angle E \quad\left(\right.$ each $\left.=40^{\circ}\right)$
$\frac{A B}{E D}=\frac{A C}{E F}$
$\therefore \triangle ABC \sim \triangle EDF$ (SAS criterion)
$\therefore \angle C=\angle F=65^{\circ}$
and $\angle B =\angle D$
But $\angle A +\angle B +\angle C =180^{\circ}$ (Sum of angles of a triangle)
$\Rightarrow 40^{\circ}+65^{\circ}+\angle C=180^{\circ}$
$\Rightarrow 105^{\circ}+\angle C=180^{\circ}$
$\therefore \angle C=180^{\circ}-105^{\circ}=75^{\circ}$

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