Question 14 Marks
Read the following text carefully and answer the questions that follow:
As shown In the village of Surya there was a big pole $PC$. This pole was tied with a strong wire of $10 m$ length. Once there was a big spark on this pole, thus wires got damaged very badly. Any small fault was usually repaired with the help of a rope which normal board electricians were carrying on bicycles.
This time electricians need a staircase of $10 m$ so that it can reach at point $P$ on the pole and this should make $60^{\circ}$ with line $AC. $
$i.$ Show that $\triangle APC$ and $\triangle BPC$ are congruent.
$ii.$ Find the value of $\angle x$.
$iii.$ What is the value of $\angle PBC$ ?
OR
Find the value of $\angle y$.
As shown In the village of Surya there was a big pole $PC$. This pole was tied with a strong wire of $10 m$ length. Once there was a big spark on this pole, thus wires got damaged very badly. Any small fault was usually repaired with the help of a rope which normal board electricians were carrying on bicycles.
This time electricians need a staircase of $10 m$ so that it can reach at point $P$ on the pole and this should make $60^{\circ}$ with line $AC. $
$i.$ Show that $\triangle APC$ and $\triangle BPC$ are congruent.
$ii.$ Find the value of $\angle x$.
$iii.$ What is the value of $\angle PBC$ ?
OR
Find the value of $\angle y$.
Answer
View full question & answer→$\text { i. In } \triangle APC \text { and } \triangle BPC$
$AP = BP \text { (Given) }$
$CP = CP \text { (common side) }$
$\angle ACP =\angle BCP =90^{\circ}$
By RHS criteria $\triangle APC \cong \triangle BPC$
$ii.$ In $\triangle ACP , APC ^{+} \angle PAC +\angle ACP =180^{\circ}$
$\Rightarrow x+60^{\circ}+90^{\circ}=180^{\circ}(\text { angle sum property of } \triangle)$
$\Rightarrow \angle x=180^{\circ}-150^{\circ}=30^{\circ}$
$\angle x=30^{\circ}$
$iii.$ In $\triangle APC$ and $\triangle BPC$
Corresponding part of congruent triangle
$\angle PAC =\angle PBC$
$\Rightarrow \angle PBC =60^{\circ} \text { (given } \angle PAC =60^{\circ} \text { ) }$
$\#\#\#$ In $\triangle APC$ and $\triangle BPC$
Corresponding part of congruent triangle
$\angle X =\angle Y$
$\Rightarrow \angle Y =30^{\circ}\left(\text { given } \angle X =30^{\circ}\right)$
$AP = BP \text { (Given) }$
$CP = CP \text { (common side) }$
$\angle ACP =\angle BCP =90^{\circ}$
By RHS criteria $\triangle APC \cong \triangle BPC$
$ii.$ In $\triangle ACP , APC ^{+} \angle PAC +\angle ACP =180^{\circ}$
$\Rightarrow x+60^{\circ}+90^{\circ}=180^{\circ}(\text { angle sum property of } \triangle)$
$\Rightarrow \angle x=180^{\circ}-150^{\circ}=30^{\circ}$
$\angle x=30^{\circ}$
$iii.$ In $\triangle APC$ and $\triangle BPC$
Corresponding part of congruent triangle
$\angle PAC =\angle PBC$
$\Rightarrow \angle PBC =60^{\circ} \text { (given } \angle PAC =60^{\circ} \text { ) }$
$\#\#\#$ In $\triangle APC$ and $\triangle BPC$
Corresponding part of congruent triangle
$\angle X =\angle Y$
$\Rightarrow \angle Y =30^{\circ}\left(\text { given } \angle X =30^{\circ}\right)$
