Question

if tan theta =5/12 then what is the value of 3sin theta+2cos theta
[tex]3sin theta + 2cos theta

Answer 1

Answer:

3

Explanation:

As per the given information in the given question, we have :

$\longmapsto \bf { \tan \; \theta = \dfrac{5}{12}}$

As we know that,

$\longmapsto \bf { \tan \; \theta = \dfrac{Perpendicular}{Base}}$

So, from two equations, we can say that :

• Perpendicular = 5
• Base = 12

We have to calculate the value of (3sin θ + 2cos θ)

In order to find that so, we need to calculate the value or sin θ and cos θ.

$\longmapsto \bf { \sin \; \theta = \dfrac{Perpendicular}{Hypotenuse}}$

$\longmapsto \bf { \cos \; \theta = \dfrac{Base}{Hypotenuse}}$

Calculating Hypotenuse :

$\longmapsto \rm { H^2 = B^2 + P^2}$

$\longmapsto \rm { H^2 = (12)^2 + (5)^2}$

$\longmapsto \rm { H^2 =144+25}$

$\longmapsto \rm { H^2 = 169}$

$\longmapsto \rm { H = \sqrt{169}}$

$\longmapsto \bf { H=13}$

Therefore, hypotenuse is 13.

Finding the value of sin θ :

$\longmapsto \bf { \sin \; \theta = \dfrac{Perpendicular}{Hypotenuse}}$

$\longmapsto \rm { \sin \; \theta = \dfrac{5}{13}}$

Finding the value of cos θ :

$\longmapsto \bf { \cos \; \theta = \dfrac{Base}{Hypotenuse}}$

$\longmapsto \rm { \cos \; \theta = \dfrac{12}{13}}$

$\rule{200}2$

Now, solving the expression we are given.

$\longmapsto \bf { 3\sin \; \theta + 2\cos \; \theta}$

Substituting the value of sin θ and cos θ.

$\longmapsto \rm { 3\Bigg ( \dfrac{5}{13}\Bigg )+ 2\Bigg ( \dfrac{12}{13} \Bigg )}$

$\longmapsto \rm { \dfrac{15}{13} + \dfrac{24}{13} }$

$\longmapsto \rm { \dfrac{15+24}{13} }$

$\longmapsto \rm { \dfrac{39}{13} }$

$\longmapsto \bf { 3}$

∴ The required value is 3.

Answer 2

Answer:

+3

Explanation:

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

=> (12)^2 + (5)^2

=> 144+25

=> √169

=> 13

sin = opposite / hypotenuse

cos = adjacent/ hypotenuse

=> 3(5/13) + 2(12/13)

=> 15/13 + 24/13

=> 39/13

=> +3