Question

In Fig. 6.42, if lines PQ and RS intersect at point T, such that PRT = 40°, RPT = 95° and TSQ = 75°, find SQT.​

Answer 1

$\large \star \underline{ \underline{ \sf \: given : }}$

• PRT = 40°

• RPT = 95°

• TSQ = 75°

$\\ \large \star \underline{ \underline{ \sf \: to \: \: find: }}$

• SQT

$\\ \large \star \underline{ { \sf \: solving: }}$

$\: \: \: \: \: \: \: \: \: \: \: \large \star \underline{\tt \: angle \: \: sum \: \: property \:} \star$

$\large \sf \: sum \: \: of \: \: 3 \: angles = 180 \degree$

$\large \sf: \implies \: \angle PRT + \angle RPT + \angle PTR = 180 \degree \\ \\ \large \sf: \implies \: 40 \degree+ 95 \degree + \angle PTR = 180 \degree \\ \\ \large \sf: \implies \: 135 \degree + \angle PTR = 180 \degree \\ \\ \large \rm \mapsto \: \: transposing \: \: terms : \\ \\ \large \sf: \implies \: \angle PTR = 180 \degree - 135 \degree \\ \\ \large \sf \therefore \: \: \underline{ \angle PTR = 45\degree }$

$\:$

$\large \rightarrow \: \sf \angle PTR = \angle STQ \\ \large \sf \: \: (vertically \: \: opposite \: \: angle) \\ \\ \large : \implies \: \sf \: 45 \degree = 45 \degree$

$\: \: \: \: \: \: \: \: \: \: \: \large \star \underline{\tt \: angle \: \: sum \: \: property \:} \star$

$\large \sf : \implies \: \angle SQT + \angle STQ + \angle TSQ = 180 \degree \\ \\ \large \sf : \implies \: \angle SQT +45 \degree + 75 \degree = 180 \degree \\ \\ \large \sf : \implies \: \angle SQT +120 \degree= 180 \degree \\ \\ \large \sf : \implies \: \angle SQT = 180 \degree - 120 \degree$

$\: \: \large \boxed{ \underline{ \sf \therefore \: \: \angle SQT = 60 \degree}}$