Question

Anyone please help me to solve no. 20

Answer 1

Hey......!!! here is ur answer......☺️☺️☺️

Given that,

5cosA–12sinA = 0

=> 5cosA = 12sinA

=> sinA/cosA = 5/12

=> tanA = 5/12 = P/B

{ where Perpendicular = 5 & Base = 12 }

then according to Pythagoras theoram....

H² = P²+B²

=> H² = 5²+12²

=> H² = 25+144 = 169

=> H = 13 then Hypotenuse = 13

Now, sinA = P/H = 5/13 and cosA = B/H = 12/13

Then we have to find...

(sinA+cosA)/(2cosA–sinA)

=> (5/13+12/13)/(2×12/13–5/13)

=> (17/13)/(24/13–5/13)

=> (17/13)/(24–5)/13

=> 17/13/19/13

=> 17/19

=> (sinA+cosA)/(2cosA–sinA) = 17/19.

I hope it will be helpful for you.....✌️✌️✌️

Answer 2

hi dear I don't no the answer but can we talk for a while