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Triangles — Maths STD 9 — Question

Gujarat BoardEnglish MediumSTD 9MathsTriangles1 Mark
MCQ
In the given figure, $\text{EAD}\perp\text{BCD}.$ Ray $FAC$ cuts ray $EAD$ at a point A such that $\angle\text{EAF}=30^\circ.$ Also, in $\triangle\text{BAC},\angle\text{BAC}=\text{x}^\circ$ and $\angle\text{ABC}=(\text{x}+10).$ Then, value of $x$ is:
  • A
    $20$
  • ✓
    $25$
  • C
    $30$
  • D
    $35$
✓

Answer

Correct option: B.
$25$
$\angle\text{EAF}=\angle\text{CAD}$ (vertically opposite angles)
$\Rightarrow\angle\text{CAD}=30^\circ$
In $\triangle\text{ABD},$ by angle sum property
$\angle\text{A}+\angle\text{B}+\angle\text{D}=180^\circ$
$\Rightarrow(\text{x}+30)^\circ+(\text{x}+10)^\circ+90^\circ=180^\circ$
$\Rightarrow2\text{x}+130^\circ=180^\circ$
$\Rightarrow2\text{x}=50^\circ$
$\Rightarrow\text{x}=25^\circ$

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