Question

sinq=4/3,find the value of 3sinq+5cosq/5sinq-3cosq​

Answer 1

Answer:

27/11

Step-by-step explanation:

Correct Question :-

tanQ = 4/3 , find the value of '3sinQ + 5cosQ/5sinQ - 3cosQ'.

Given,

tanQ =4/3

To Find :-

Value of :-

$\dfrac{3sinQ+5cosQ}{5sinQ-3cosQ}$

How To Do :-

As they given the value of 'tanQ' By using pythagoras theorem we need find the value of hypotenuse side and then we need to find the value of 'sinQ and cosQ' by using the ratios formula of trigonometry and we need to substitute those values.

Formula Required :-

$tan\theta=\dfrac{opposite\:side}{adjacent\:side}$

$sin\theta=\dfrac{opposite\:side}{hypotenuse\:side}$

$cos\theta=\dfrac{adjacent\:side}{hypotenuse\:side}$

Pythogoras theorem :-

(hypotenuse side)² = (opposite side)² + (adjacent side)²

Solution :-

$tanQ=\dfrac{4}{3}=\dfrac{opposite\:side}{adjacent\:side}$

→ Opposite side = 4 , Adjacent side = 3

Let , hypotenuse side be 'x'

Applying pythagoras theorem :-

x² = 4² + 3²

x² = 16 + 9

x² = 25

Applying square root on both sides :-

x = 5

∴ Hypotenuse Side = x = 5

$sinQ=\dfrac{opposite\:site}{hypotenuse}$

= 4/5

∴ sinQ = 4/5

$cosQ=\dfrac{adjacent\:side}{hypotenuse\:side}$

= 3/5

∴ cosQ = 3/5

$\dfrac{3sinQ+5cosQ}{5sinQ-3cosQ}$

$=\dfrac{\left(3\times \dfrac{4}{5}\right)+\left(5\times \dfrac{3}{5}\right)}{\left(5\times \dfrac{4}{5}\right)-\left(3\times\dfrac{3}{5}\right)}$

$=\dfrac{\dfrac{12}{5}+\dfrac{15}{5}}{\dfrac{20}{5}-\dfrac{9}{5}}$

$=\dfrac{\dfrac{27}{5}}{\dfrac{11}{5}}$

$=\dfrac{27}{5}\times \dfrac{5}{11}$

$=\dfrac{27}{11}$

$\therefore\dfrac{3sinQ+5cosQ}{5sinQ-3cosQ}=\dfrac{27}{11}$