Question
In a $\triangle\text{ABC},\angle\text{B}=90^\circ,\angle\text{AB}=12\text{cm}$ and BC = 5cm.
Find:
  1. $\cos\text{A}$
  2. $\text{cosec}\text{A}$
  3. $\cos\text{C}$
  4. $\text{cosec}\text{C}.$

Answer


In $\triangle\text{ABC},\angle\text{B}=90^\circ$
AB = 12cm and BC = 5cm
By Pythagoras theorem, we have
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$=12^2+5^2=144+25=169$
$\Rightarrow\text{AC}=13\text{cm}$
  1. $\cos\text{A}=\frac{\text{Base}}{\text{Hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac{12}{13}$
  2. $\text{cosec}\text{A}=\frac{\text{Hypotenuse}}{\text{Perpendicular}}=\frac{\text{AC}}{\text{BC}}=\frac{13}{5}$
  3. $\cos\text{C}=\frac{\text{Base}}{\text{Hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{5}{13}$
  4. $\text{cosec}\text{C}=\frac{\text{Hypotenuse}}{\text{Perpendicular}}=\frac{\text{AC}}{\text{AB}}=\frac{13}{12}$

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