Question

Given tan A =4/3. find the trigonometric ratios of the angle A. ​

Answer 1

Answer:

$\tan(a) = \frac{p}{b} \\ p = 4 \:, \: b \: = 3 \\ {h}^{2} = {p}^{2} + {b}^{2} \\ {h}^{2} = {4}^{2} + {3}^{2} \\h = \sqrt{25} \\ h = 5 \\ \sin(a) = \frac{p}{h} = \frac{4}{5} \\ \csc(a) = \frac{5}{4} \\ \cos(a) = \frac{b}{h} = \frac{3}{5} \\ \sec(a) = \frac{5}{3} \\ \cot(a) = \frac{3}{4}$

Answer 2

Topic:-

Trigonometry

Question:-

$Given\: tan\theta=4/3. find\:the\:trigonometric\: ratios\: of \:the\: angle\: \theta$

Solution:-

$Given tan \theta=4/3$

$We\: know\: that \:tan\theta =\dfrac{Opposite\:side}{Adjacent\:side}$

$Here, \:Opposite\:side= 4 \: and Adjacent\:side=3$

$We \:need \:Hypotenuse\:side \: here$

$We\: know\: that\:Pythageorous\:theorem$

$Hypotenuse²=Opposite\:side²+Adjacent\:side²$

$Hypotenuse²= 4²+3²$

$Hypotenuse²= 16+9$

$Hypotenuse²= 25$

$Hypotenuse=\sqrt{25}$

$Hypotenuse= 5$

$Now\:we\:got\:all\:sides$

$Sin\theta =\dfrac{Opposite\:side}{Hypotenuse\:side}$

So, $Sin\theta=\dfrac{4}{5}$

$cos\theta=\dfrac{Adjacent\:side}{Hypotenuse\:side}$

$cos\theta=\dfrac{ 3}{5}$

$cot\theta=\dfrac{Adjacent\:side}{Opposite\:side}$

$cot\theta=\dfrac{3}{4}$

$csc\theta=\dfrac{Hypotenuse\:side}{Opposite\:side}$

$csc\theta=\dfrac{5}{4}$

$sec\theta= \dfrac{Hypotenuse}{Adjacent}$

$sec\theta= \dfrac{ 5}{3}$

More Information:-

Trigon metric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

csc²θ - cot²θ = 1

Trigometric relations

sinθ = 1/cscθ

cosθ = 1 /secθ

tanθ = 1/cotθ

tanθ = sinθ/cosθ

cotθ = cosθ/sinθ

Trigonmetric ratios

sinθ = opp/hyp

cosθ = adj/hyp

tanθ = opp/adj

cotθ = adj/opp

cscθ = hyp/opp

secθ = hyp/adj