Question

If \[\dfrac{{(\cos {\text{ A)}}}}{3} = \dfrac{{(\cos {\text{ B)}}}}{4} = \dfrac{1}{5}\] , \[ - \dfrac{\pi }{2} < A < 0\],\[ - \dfrac{\pi }{2} < B < 0\], then value of \[2\sin A + 4\sin B\] is ?
A. \[4\]
B. \[{\text{ - 2}}\]
C. \[{\text{ - 4}}\]
D. \[{\text{0}}\]

Answer 1

Hint: Whenever we have trigonometric ratios given with some values equal to them always try to consider a triangle and use trigonometric ratios formula to find the required side of the triangle such as in \[\cos \]we have \[\dfrac{{base}}{{hypotenuse}}\], similarly this applied for another trigonometric ratios .
Complete step by step answer:
Consider a right angled triangle ABC ,

Given :
 \[\dfrac{{(\cos {\text{ A)}}}}{3} = \dfrac{{(\cos {\text{ B)}}}}{4} = \dfrac{1}{5}\], first we take the terms ,
\[\dfrac{{(\cos {\text{ A)}}}}{3} = \dfrac{1}{5}\] , on solving it we get ,
\[\cos {\text{ A}} = \dfrac{3}{5}\] , on comparing with trigonometric ratios of cosine \[ = \dfrac{{base}}{{hypotenuse}}\] , we have
\[base = 3\] and \[hypotenuse = 5\] , therefore

In the question we have to solve for \[\sin \], so trigonometric ratios for sine is \[\dfrac{{perpendicular}}{{hypotenuse}}\], so for perpendicular , we will use Pythagoras theorem , therefore we have \[A{B^2} + B{C^2} = A{C^2}\] ,
Putting the values of \[AB\] and \[AC\] we have ,
\[B{C^2} + {3^2} = {5^2}\]
\[B{C^2} = {5^2} - {3^2}\]
On solving we get
\[B{C^2} = 25 - 9\]
\[B{C^2} = 16\]
Now taking square root on both the sides we have ,
\[\sqrt {B{C^2}} = \sqrt {16} \]
\[BC = 4\]
For \[\sin A\]we have \[\dfrac{{BC}}{{AC}}\] , now putting the values of \[BC\] and \[AC\] , we get
\[\sin A = \dfrac{4}{5}\] .
Now , we take the terms \[\dfrac{{(\cos {\text{ B)}}}}{4} = \dfrac{1}{5}\] ,

\[\cos B = \dfrac{4}{5}\] , similarly we can find \[\sin B\] using Pythagoras theorem , we get
\[A{B^2} + A{C^2} = B{C^2}\]
Putting the values we have
\[{4^2} + A{C^2} = {5^2}\]
\[A{C^2} = {5^2} - {4^2}\]
On solving further we get
\[A{C^2} = 16\]
Taking square root on both the sides
\[\sqrt {A{C^2}} = \sqrt 9 \]
\[AC = 3\]
For \[\sin {\text{ B}}\]we have \[\dfrac{{AC}}{{BC}}\] , now putting the values of \[BC\] and \[AC\] , we get
\[\sin {\text{ B = }}\dfrac{3}{5}\], now putting the values in the given question , we get
\[2\sin A + 4\sin B = \dfrac{8}{5} + \dfrac{{12}}{5}\] ,
\[2\sin A + 4\sin B = \dfrac{{20}}{5}\]
On further solving we get
\[2\sin A + 4\sin B = 4\].

So, the correct answer is “Option A”.

Note: Always remember the trigonometric ratios as without them questions related to it will not be solved . Try to solve using a triangle as it will help assigning the angles with corresponding sides such as base , hypotenuse and perpendicular .