Question

Solve the ques 5 in the attachment

Class 10

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Answer 1

Hey!!!!!

Good Evening

Difficulty Level : Medium

Chances of being asked in Board : 70%

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For the figure refer to the diagram

Here we are given

sinA = 3/5

Let the opposite that is BC be 3x and hypotenuse be 5x

Then by applying Pythagoras theorem we have

=> AC = 4x

Now Let's calculate we have to find

cosB = BC/AB

=> cosB = 3/5

Similarly cosA = 4/5 and sinB = 4/5

To Find : sinAcosB + cosAsinB

$= ( \frac{3}{5} )(\frac{3}{5} ) + ( \frac{4}{5} )( \frac{4}{5} )$

Thus

$= \frac{9}{25} + \frac{16}{25}$

= 25/25

= 1 <<<<<< Answer

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Hope this helps ✌️

Good Night

Jai Hind

Answer 2

Hey mate!


Here is yr answer.....


Given,

sin A = 3/5

Sin A = opp. side/Hyp = BC/AC

Let .. BC = 3k, AC = 5k (where k is any real no.)


By pythagorous theorem ,

AC² = AB² - BC²

AC² = (5k)² - (3k)²

AC² = 25k² - 9k²

AC² = 16k²

AC = √16k²

AC = 4k


Now,

sin B = opp. side/Hyp

sin B = AC/AB

sin B = 4k/5k

sin B = 4/5

_____________________

cos A = Adj. Side/Hyp

Cos A = AC/AB

cos A = 4k/5k

cos A = 4/5

___________________

cos B = BC/AB

cos B = 3k/5k

cos B = 3/5


So,

sin A = 3/5 , Sin B = 4/5

cos A = 4/5 , cos B = 3/5


=> sin A cos B + cos A sin B

=> 3/5 × 3/5 + 4/5 × 4/5

=> 9/25 + 16/25

=> 9+16/25

=> 25/25

=> 1


Hope it helps u...


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