Question

If cot theta =8/15 then find the other trigonometric ratios​

Answer 1

We have:

$\cot \theta$ = $\dfrac{8}{15}$

If $\cot \theta$ = $\dfrac{8}{15}$, then find the other trigonometric ratios​.

Solution:

∴ $\cot \theta$ = $\dfrac{8}{15}=\dfrac{b}{p}$

Where,, b = base and p = perpendicular

∴ Hypotenuse, h = $\sqrt{p^2+b^2}$

= $\sqrt{15^2+8^2} =\sqrt{225+64} =\sqrt{289} =17$

The other trigonometric ratios​ are:

$\sin \theta$ = $\dfrac{p}{h}=\dfrac{15}{17}$

$\cos \theta$ = $\dfrac{b}{h}=\dfrac{8}{17}$

$\tan \theta$ = $\dfrac{p}{b}=\dfrac{15}{8}$

$\csc \theta$ = $\dfrac{h}{p}=\dfrac{17}{15}$

$\sec \theta$ = $\dfrac{h}{b}=\dfrac{17}{8}$

Answer 2

Given:

cot theta =8/15

To find:

The other trigonometric ratios​

Solution:

From the trigonometric ratios we know that:

Cotθ = B/P

By the pythagoras theorem;

The length of the Hypotenuse is given by:

H² = P² + B²

H² = 15² + 8²

H² = 225 + 64

H² = 289

H = 17

Other ratios is given as:

Sinθ = P/H = 15/17

Cosθ = B/H = 8/17

Tanθ = P/B = 15/8

Cosecθ = H/P = 17/15

Secθ = H/B = 17/8