An equilateral \[\Delta TQR\]is drawn inside a square PQRS. The value of the angle PTS, in degrees) is |
[SSC (10+2) 2012] |
A) 120
B) 150
C) 75
D) 90
Correct Answer: B
Solution :
Since, \[\Delta TQR\]is an equilateral triangle. |
\[\therefore \]\[\angle TQR=\angle TRQ=\angle RTQ=60{}^\circ \] |
and \[TQ=TR=RQ\] |
and \[SR=RQ=PQ=SP\][sides of square] |
Now, in \[\Delta RTS\] |
\[RT=SR\] |
\[\therefore \] \[\angle STR=RST\] |
(angle opposite to equal sides are equal] |
\[=\frac{180{}^\circ -30{}^\circ }{2}=75{}^\circ \] |
Similarly, in \[\Delta QTP,\]\[QT=PQ\] |
\[\angle PTQ=\angle QPT=75{}^\circ \] |
Now, at point T |
\[\angle PTS+\angle PTQ+\angle QTR+\angle RTS=360{}^\circ \] |
[since, angles around on point is equal to \[360{}^\circ \]] |
\[\Rightarrow \]\[\angle PTS=360{}^\circ -\,\,(75{}^\circ +60{}^\circ +75{}^\circ )\] |
\[=360{}^\circ -210{}^\circ =150{}^\circ \] |
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