A) 300 cm
B) 350 cm
C) 320 cm
D) 360 cm
Correct Answer: A
Solution :
In rt. \[\Delta \,ADQ,\frac{AD}{AQ}=\sin {{30}^{o}}\Rightarrow \frac{AD}{AQ}=\frac{1}{2}\] \[\Rightarrow \] \[AQ=2AD\Rightarrow AQ=2\times 60=120cm.\] ?.(1) In rt. \[\Delta \,ABP,\frac{AB}{AP}=\cos {{60}^{o}}\Rightarrow \frac{AB}{AP}=\frac{1}{2}\] \[\Rightarrow \] \[AP=2AB\] \[\Rightarrow \] \[AP=2\times 90\]\[=180\,cm\] ?.(2) \[\therefore \] \[AP+AQ\] \[=120+180\] \[=300\,cm.\] [From (1) and (2)]You need to login to perform this action.
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