Height and Distance Questions Answers Formulas, Tips and Tricks

• 1. Basic formulae of trigonometry used in Heights and Distances Questions

  
  

  In a right angle triangle,
  \begin{aligned}
  sin\theta = \frac{Perpendicular}{Hypotenuse}\\
  cos\theta = \frac{Base}{Hypotenuse}\\
  tan\theta = \frac{Perpendicular}{Base}\\
  cosec\theta = \frac{Hypotenuse}{Perpendicular}\\
  sec\theta = \frac{Hypotenuse}{Base}\\
  cot\theta = \frac{Base}{Perpendicular}\\
  
  \end{aligned}
  
  In school we were use to taught a simple formula to learn it, lets refresh that.
  
  Do you remember, "Pandit Badri Prasad Hari Hari Bole".
  

  Sin	Cos	Tan
  "P"andit	"B"adri	"P"rasad
  "H"ari	"H"ari	"B"ole

  
  
  So taking the initials below that sin, cos and tan, we can derive their values.
  
  cosec is simply reciprocal to sin,
  sec is reciprocal to cos,
  cot is reciprocal to tan.

  • Comment on this formula

• 2. Formulae of trigonometry

  
  

  \begin{aligned}
  sin^2\theta + cos^2 \theta = 1 \\
  1 + tan^2 \theta = sec^2 \theta \\
  1 + cot^2 \theta = cosec^2 \theta \\
  \end{aligned}

  • Comment on this formula

• 3. Trigonometric values table

  
  

  \begin{aligned}\theta\end{aligned}	\begin{aligned}0^{\circ}\end{aligned}	\begin{aligned}30^{\circ}\end{aligned}	\begin{aligned}45^{\circ}\end{aligned}	\begin{aligned}60^{\circ}\end{aligned}	\begin{aligned}90^{\circ}\end{aligned}
  sin\begin{aligned}\theta\end{aligned}	0	\begin{aligned}\frac{1}{2}\end{aligned}	\begin{aligned}\frac{1}{\sqrt{2}}\end{aligned}	\begin{aligned}\frac{\sqrt{3}}{2}\end{aligned}	\begin{aligned}1\end{aligned}
  cos\begin{aligned}\theta\end{aligned}	1	\begin{aligned}\frac{\sqrt{3}}{2}\end{aligned}	\begin{aligned}\frac{1}{\sqrt{2}}\end{aligned}	\begin{aligned}\frac{1}{2}\end{aligned}	\begin{aligned}0\end{aligned}
  tan\begin{aligned}\theta\end{aligned}	0	\begin{aligned}\frac{1}{\sqrt{3}}\end{aligned}	1	\begin{aligned}\sqrt{3}\end{aligned}	not defined

  • Comment on this formula
Height and Distance Questions Answers