Height and Distance Problems and Solutions 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.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}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