Question

If A + B, A are acute angles such that sin (A + B) = $\frac{24}{25}$ and tan A = $\frac{3}{4}$ then find the value of cos B.

Answer 1

given, A + B, A are acute angles such that sin (A + B) = $\frac{24}{25}$ and tan A = $\frac{3}{4}$.

tanA = 3/4 = p/b

so, p = 3 and b = 4

then, h = √(p²+b²) = √(3² + 4²) = 5

sinA = 3/5 and cosA = 4/5


sin(A + B) = 24/25

we know, sin(X + Y) = sinX. cosY + cosX . sinY

so, sinA. cosB + cosA. sinB = 24/25

3/5 × cosB + 4/5 × sinB = 24/25

3cosB + 4sinB = 24/5

15cosB + 20sinB = 24

dividing by cosB both sides,

15 + 20tanB = 24secB

squaring both sides,

225 + 400tan²B + 2600tanB= 576sec²B

we know, sec²x = 1 + tan²x

so, 225 + 400tan²B + 600tanB = 576 + 576tan²B

176tan²B - 600tanB + 576 - 225 = 0

176tan²B - 600tanB + 351 = 0

tanB = 3/4 , 117/44

cosB = 4/5 , 44/125

Answer 2

HELLO DEAR,




GIVEN:-
If A + B, A are acute angles such that sin (A + B) = $\frac{24}{25}$ and tan A = $\frac{3}{4}$


tanA = 3/4 = p/b
h = √(9 + 16) = √25 = 5

sinA = 3/5 , cosA = 4/5


now, sin(A + B) = sinAcosB + cosAsinB = 24/25

=> (3/5)cosB + (4/5)sinB = 24/25

=> (3cosB + 4sinB) = 24/5

=> 15cosB + 20sinB = 24

divide by cosB on both side,

=> 15 + 20tanB = 24secB

on squaring both side,

=> 225 + 400tan²B + 600tanB = 576(1 + tan²B)

=> 400tan²B + 600tanB - 576tan²B = 576 - 225

=> -176tan²B + 600tanB = 351

=> 176tan²B - 600tanB + 351 = 0

=> 176tan²B - 468tanB - 132tanB + 351 = 0

=> (4tanB - 3) (44tanB - 117) = 0

=> tanB = 3/4 , tanB = 117/44 

cosB = 4/5 , cosB = 44/125 




I HOPE IT'S HELP YOU DEAR,
THANKS