Question

In $\Delta PQR;PQ=PRA$ is a point in PQ and B is a point in PR, so that $QR=RA=AB=BP$ Find the value of $\angle Q.$

• A. ${\left(25\frac{1}{7}\right)}^{\circ }$
• B. ${\left(77\frac{1}{7}\right)}^{\circ }$
• C. ${\left(50\frac{1}{7}\right)}^{\circ }$
• D. ${\left(\frac{180}{7}\right)}^{\circ }$

Answer 1

Answer: (B)

${\left(77\frac{1}{7}\right)}^{\circ }$

Given: $△PQR$, $PQ=PR$, A and B are points such that, $QR=RA=AB=BP$
Now, In $△PAB$
$AB=PB$
Thus, $\angle P=\angle PAB=P$ (isosceles triangle property)
and $\angle ABR=\angle P+\angle PAB$
$\angle ABR=2P$ (Exterior angle property)
Now, In $△ABR$
$AB=RA$
$\angle ABR=\angle ARB=2P$ (Isosceles triangle property)
Sum of angles of the triangle = 180
$\angle ABR+\angle ARB+\angle BAR=180$
$2P+2P+\angle BAR=180$
$\angle BAR=180-4P$
Sum of angles on a straight line = 180
$\angle PAB+\angle BAR+\angle RAB=180$
$P+180-4P+\angle RAB=180$
$\angle RAB=3P$
In $△QRA$
$QR=RA$
$\angle Q=\angle QAR=3P$ (Isosceles triangle property)
But , $PQ=PR$
$\angle Q=\angle R=3P$
Now, Sum of angles of $△PQR$ = 180
$\angle P+\angle Q+\angle R=180$
$P+3P+3P=180$
$P=\frac{180}{7}$
hence, $\angle Q=\frac{3\times 180}{7}=77{\frac{1}{7}}^{\circ }$