| The angles of a triangle are in arithmetic progression. The ratio of the least angle (in degrees) to the number of radians in the greatest angle is \[60:\text{ }\pi \]. The angles (in degrees) are [SSC (CGL) 2012] |
A) \[30{}^\circ ,\]\[60{}^\circ ,\]\[90{}^\circ \]
B) \[35{}^\circ ,\]\[55{}^\circ ,\]\[90{}^\circ \]
C) \[40{}^\circ ,\]\[50{}^\circ ,\]\[90{}^\circ \]
D) \[40{}^\circ ,\]\[55{}^\circ ,\]\[85{}^\circ \]
Correct Answer: A
Solution :
| (a) |
| Let angles of a triangle in AP be \[(a-d){}^\circ ,\]\[a{}^\circ ,\]\[(a+d){}^\circ .\] |
| \[\therefore \] \[a-d+a+a+d=180{}^\circ \] [since, sum of all angles of triangle is \[180{}^\circ \]] |
| \[\Rightarrow \]\[3a=180{}^\circ \]\[\Rightarrow \]\[a=60{}^\circ \] |
| Now, given ratio of least angle to largest angle is \[60:\pi ,\] then |
| \[\therefore \] \[\frac{a-d}{a+d}=\frac{60{}^\circ }{\pi }=\frac{60{}^\circ }{180{}^\circ }=\frac{1}{3}\]\[\{\pi =180{}^\circ \}\] |
| \[\Rightarrow \] \[\frac{60{}^\circ -d}{60{}^\circ +d}=\frac{1}{3}\] \[\Rightarrow \] \[180{}^\circ -3d=60{}^\circ +d\] |
| \[\Rightarrow \] \[4d=120{}^\circ \] \[\Rightarrow \] \[d=30{}^\circ \] |
| \[\therefore \]Angles of triangle are \[a-d=60{}^\circ -30{}^\circ =30{}^\circ \] |
| \[\Rightarrow \] \[a=60{}^\circ \] and \[a+d=60{}^\circ +30{}^\circ =90{}^\circ \] |
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