Question

In a right $△$ ABC, if $\angle$ A is acute and tanA = 3/4, find the remaining trigonometric ratios of $\angle$ A.

Answer 1

Consider a $\Delta$ ABC in which $\angle$B = ${90}^{\circ }$

For $\angle$A , we have :

Base = AB, perpendicular = BC and hypotenuse = AC

∴ tan A = $\frac{Perpendicular}{Base}$ = $\frac{3}{4}$

⇒ $\frac{BC}{AB}=\frac{3}{4}$

Let, BC = 3x units and AB = 4x units.

Then, AC = $\sqrt{A{B}^2+B{C}^2}$

= $\sqrt{(4x{)}^2+(3x{)}^2}$

= $\sqrt{\left(25x^2\right)}$ = 5x units.

Now, SinA = opposite/Hypotenuse

SinA = 3/5

Similarly , CosA = 4/5

Hence , CosecA = 5/3 ; SecA = 5/4 ; CotA = 4/3