Q.

From the following options, find the value of cosAsinB+sinAcosB,   if sinA= 4 5   and cosB= 12 13  .


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a

3 5  

b

1

c

63 65  

d

3 65   

answer is C.

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Detailed Solution

Given that, sinA= 4 5   and cosB= 12 13  .
Write sinA= 4 5   in terms of the ratio of sides of the right triangle.
sinA= 4 5   4k 5k = Opposite side Hypotenuse  
With respect to A  , form a triangle with opposite side (BC) as 4k and the hypotenuse (AC) as 5k.
Question ImageFind third side AB   by applying Pythagoras theorem.
(base) 2 + (perpendicular) 2 = (hypotenuse) 2  
A C 2 =B C 2 +A B 2 25 k 2 =16 k 2 +A B 2 A B 2 =25 k 2 16 k 2 AB=3k   A  
Write cosA   from sides of the triangle.
cosθ= Adjacent side Hypotenuse  
cosA= AB AC cosA= 3k 5k cosA= 3 5  
Write cosB= 12 13   in terms of the ratio of sides of the right triangle.
  cosB= 12 13   12k 13k = Adjacent side Hypotenuse  
We know that, cosθ= Adjacent side Hypotenuse  .
With respect to B, form a triangle with adjacent side (AB) as 12k and the hypotenuse (BC) as 13k.
Question ImageFind third side AC   by applying Pythagoras theorem.
(base) 2 + (perpendicular) 2 = (hypotenuse) 2  
B C 2 =A C 2 +A B 2 169 k 2 =A C 2 +144 k 2 A C 2 =169 k 2 144 k 2 A C 2 =25 k 2 AC=5k  
Write sinB   from sides of the triangle.
sinθ= Opposite side Hypotenuse   sinB= AC BC sinB= 5k 13k sinB= 5 13   To find the value of expression put the values of sinB   and cosA  . Question Image Thus, cos A sin B+ sin A cos B= 63 65  .
Hence, option 3) is correct.
 
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