Mathematics

In the figure $\angle P + \angle Q + \angle R + \angle S + \angle T = 2$ right angles

True

SOLUTION
Sum of angles on a straight line=${ 180 }^{ \circ }$
$\implies\quad \angle PAE={ 180 }^{ \circ }-\angle 1$
$Similarly,\angle PEA={ 180 }^{ \circ }-\angle 5$
$In\triangle PAE$
$\angle PAE+\angle PEA+\angle APE={ 180 }^{ \circ }$
$\implies\quad 180-\angle 1+180-\angle 5+\angle APE={ 180 }^{ \circ }$
$\implies\quad \angle APE=\angle 1+\angle 5-{ 180 }^{ \circ }=\angle P$
$Similarly,\angle BSC=\angle 2+\angle 3-{ 180 }^{ \circ }=\angle S$
$\angle DRC=\angle 3+\angle 4-{ 180 }^{ \circ }=\angle R$
$\angle DQE=\angle 4+\angle \angle 5-{ 180 }^{ \circ }=\angle Q$
$\angle ATS=\angle 1+\angle 2-{ 180 }^{ \circ }=\angle T$
$\implies\quad \angle P+\angle Q+\angle R+\angle S+\angle T$
$=\angle 1+\angle 2+\angle 3+\angle 4+\angle 2+\angle 3+\angle 4+\angle 5+\angle 1+\angle 5-(180\times 5)$
$=2(\angle 1+\angle 2+\angle 3+\angle 4+\angle 5)-900\quad -(1)$
From pentagon ABCDE,
Sum of angles of pentagon=$((2\times 5)-4)\times { 90 }^{ \circ }$
($\because$ Sum of angles=$(2n-4)\times { 90 }^{ \circ }$)
$=6\times 90={ 540 }^{ \circ }$
$\implies\quad \angle P+\angle Q+\angle R+\angle S+\angle T=2({ 540 }^{ \circ })-{ 900 }^{ \circ }$
$={ 1080 }^{ \circ }-{ 900 }^{ \circ }$
$={ 180 }^{ \circ }$
$=2\times 90$
$=2\quad Right\quad angles$
$A)True$

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TRUE/FALSE Medium Published on 09th 09, 2020
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