Question

Given that sin¢ = 3/5 .

Find other trigonometry ratio if ¢ lies in first quadrant.​

Answer 1

Answer:

Your answer is in the above attachment

Answer 2

$\mathfrak{\underline{\underline{\green{Answer:-}}}}$

cosA = $\dfrac{4}{5}$

tanA = $\dfrac{3}{4}$

cotA = $\dfrac{4}{3}$

secA = $\dfrac{5}{4}$

cosecA = $\dfrac{5}{3}$

$\mathfrak{\underline{\underline{\green{Explanation:-}}}}$

Given:

sinA = $\dfrac{3}{5}$

To Find:

The other trigonometric ratios in the first quadrant

Solution:

In a right ΔABC, [refer the attachment]

$\mathsf{ sinA = \dfrac{3}{5}}$

As we know

$\boxed{\pink{ sin A = \dfrac{Opposite\: side}{Hypotenuese}}}$

So,

$\mathsf{ sinA = \dfrac{3}{5} = \dfrac{BC}{AC}}$

Therefore

BC = 3 units

AC = 5 units

AB = ??

Let us find the value of AB

From Pythagoras theorem

$\boxed{\pink{ {AC}^{2}= {AB}^{2}+{BC}^{2}}}$

$\mathsf{ {AC}^{2}= {AB}^{2}+{BC}^{2}}$

$\mathsf{ {5}^{2}= {AB}^{2}+{3}^{2}}$

$\mathsf{ 25= {AB}^{2}+{9}}$

$\mathsf{ {AB}^{2}= 25-9}$

$\mathsf{ {AB}^{2}= 16}$

$\mathsf{ AB= \sqrt{16} }$

$\mathsf{ AB= 4 \: units}$

Let us find the other ratios

CosA

$\boxed{\red{ cosA = \dfrac{Adjacent\: side}{Hypotenuese}}}$

$\mathsf{ cosA = \dfrac{AB}{AC}}$

$\mathsf{ cosA = \dfrac{4}{5}}$

tanA

$\boxed{\orange{ tanA = \dfrac{Opposite\: side}{Adjacent \: side}}}$

$\mathsf{ tanA = \dfrac{BC}{AB}}$

$\mathsf{ tanA= \dfrac{3}{4}}$

cotA

$\boxed{\pink{ cot A = \dfrac{Adjacent\: side}{Opposite \: side}}}$

$\mathsf{ cotA = \dfrac{AB}{BC}}$

$\mathsf{ cotA = \dfrac{4}{3}}$

secA

$\boxed{\green{ secA = \dfrac{Hypotenuese}{Adjacent \: side}}}$

$\mathsf{ secA = \dfrac{AC}{AB}}$

$\mathsf{ secA = \dfrac{5}{4}}$

cosecA

$\boxed{\blue{ cosecA = \dfrac{Hypoteneuse}{Opposite\: side}}}$

$\mathsf{ cosecA = \dfrac{AC}{BC}}$

$\mathsf{ cosecA = \dfrac{5}{3}}$

$\mathbb{\red{NOTE:-}}$

Assume "A" as '¢'