Question

In the given triangle $AB=3cm$ and $AC=5cm$. Find all trignometric ratios.

• A. $\sin \theta =\frac{P}{H}=\frac{4}{5}$
  $\cos \theta =\frac{B}{H}=\frac{4}{5}$
  $\tan \theta =\frac{P}{B}=\frac{4}{3}$
  $\cot \theta =\frac{B}{P}=\frac{3}{4}$
  $\sec \theta =\frac{H}{B}=\frac{5}{3}$
  $co\sec \theta =\frac{H}{P}=\frac{5}{4}$
• B. $\sin \theta =\frac{P}{H}=\frac{4}{5}$
  $\cos \theta =\frac{B}{H}=\frac{2}{5}$
  $\tan \theta =\frac{P}{B}=\frac{4}{3}$
  $\cot \theta =\frac{B}{P}=\frac{3}{4}$
  $\sec \theta =\frac{H}{B}=\frac{5}{3}$
  $co\sec \theta =\frac{H}{P}=\frac{5}{4}$
• C. $\sin \theta =\frac{P}{H}=\frac{4}{5}$
  $\cos \theta =\frac{B}{H}=\frac{1}{5}$
  $\tan \theta =\frac{P}{B}=\frac{4}{3}$
  $\cot \theta =\frac{B}{P}=\frac{3}{4}$
  $\sec \theta =\frac{H}{B}=\frac{5}{3}$
  $co\sec \theta =\frac{H}{P}=\frac{5}{4}$
• D. $\sin \theta =\frac{P}{H}=\frac{4}{5}$
  $\cos \theta =\frac{B}{H}=\frac{3}{5}$
  $\tan \theta =\frac{P}{B}=\frac{4}{3}$
  $\cot \theta =\frac{B}{P}=\frac{3}{4}$
  $\sec \theta =\frac{H}{B}=\frac{5}{3}$
  $co\sec \theta =\frac{H}{P}=\frac{5}{4}$

Answer 1

The correct option is D $\sin \theta =\frac{P}{H}=\frac{4}{5}$

$\cos \theta =\frac{B}{H}=\frac{3}{5}$
$\tan \theta =\frac{P}{B}=\frac{4}{3}$
$\cot \theta =\frac{B}{P}=\frac{3}{4}$
$\sec \theta =\frac{H}{B}=\frac{5}{3}$
$co\sec \theta =\frac{H}{P}=\frac{5}{4}$
Using pythogorus theorem
${AC}^2={AB}^2+{BC}^2$
$\Rightarrow$ $5^2=3^2+P^2$
$\Rightarrow$ $P^2=16$
$\Rightarrow$ $P=4cm$
Here, $P=4cm,B=3cm.H=5cm$
$\therefore$ $\sin \theta =\frac{P}{H}=\frac{4}{5}$
$\cos \theta =\frac{B}{H}=\frac{3}{5}$
$\tan \theta =\frac{P}{B}=\frac{4}{3}$
$\cot \theta =\frac{B}{P}=\frac{3}{4}$
$\sec \theta =\frac{H}{B}=\frac{5}{3}$
$co\sec \theta =\frac{H}{P}=\frac{5}{4}$