A) \[\frac{2\sqrt{8}}{15}\]
B) \[\frac{8}{15}\]
C) \[\frac{\sqrt{2}}{17}\]
D) \[\frac{8\sqrt{2}}{17}\]
Correct Answer: B
Solution :
[b] Given,\[\cos \theta =\frac{15}{17}\] \[\therefore \]Perpendicular \[(AB)=\sqrt{{{17}^{2}}-{{15}^{2}}}\] \[=\sqrt{289-225}\] \[=\sqrt{64=8}\] \[\therefore \] \[\cot \,(90{}^\circ -\theta )=tan\theta =\frac{8}{15}\] |
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