MCQ
If $15 \cot A=8$, then $\sin A =$ ?
  • A
    $\frac{8}{15}$
  • B
    $\frac{8}{17}$
  • C
    $\frac{17}{8}$
  • $\frac{15}{17}$

Answer

Correct option: D.
$\frac{15}{17}$
Image

$\therefore \tan A=\frac{15}{8}=\frac{BC}{AB}$
In $\triangle ABC$, by Pythagoras Theorem
$(AC)^2=(AB)^2+(BC)^2$
$=(8)^2+(15)^2$
$=64+225$
$=289$
$\Rightarrow AC=\sqrt{289}=17$
$\therefore \sin A=\frac{BC}{AC}=\frac{15}{17}$

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