Question

In a triangle ABC right angled at B,
$\angle ACB=\theta .$
$\text{If}tan\theta =\frac{3}{4},\text{then find}sin\theta$.

• A. $\frac{3}{5}$
• B. $\frac{4}{5}$
• C. $\frac{5}{4}$
• D. $\frac{5}{3}$

Answer 1

The correct option is A $\frac{3}{5}$


Given, triangle ABC is right angled at B,

$\text{Also,}tan\theta =\frac{3}{4}$

$tan\theta =\frac{AB}{BC}$

Let AB = 3k and BC = 4k

Applying Pythagoras theorem, we get,

$A{B}^2+B{C}^2=A{C}^2$

$A{C}^2=(3k{)}^2+(4k{)}^2$

$A{C}^2=25k^2$

$AC=5k$

$\Rightarrow AB:BC:AC=3:4:5$

$\Rightarrow sin\theta =\frac{AB}{AC}=\frac{3}{5}$