sinA=3/4cosA=5/4tanA=3/4

sinA=3/4cosA=5/4tanA=3/5

sinA=perpendicular/basep=3, h=4by pythagoras the ram hyponetious =hb=baseh^2=p^2+b^24^2=3^2+b^216=9+b^216-9=b^27=b^2take squre root of both side√7=bcosA= b/h =√7/4tanA= p/b =3/4ans

for cos root 7/4 and tan is 3root by4

sin =P/H then sin=3/4 ,[under root3 square + 4 square]=25 under root then we write normal 5 because square root of 5 is 25 then base is 5. cosA=5/4 and ranA=3/5.

sin A=3/4cosA=5/4tanA=3/4

Sin A=3/4Cos A=5/4tan A =3/4

given sinA = 3/4 = perpendicular/ hypotenuse perpendicular = 3hypotenuse = 4

we have to find base,h^2 = b^2 + p^24^2 = b^2 + 3^216 = b^2 + 916 - 9 = b^27 = b^2

base = root7

sinA = 3/4

cosA = base/ hypotenuse = root7/ 4

tanA = Perpendicular/ base

= 3 / root7

ANSWER sinA=3/4cosA=5/4tanA=3/4

sinA=3/4 लंब/कर्णcosA=5/4tanA=3/4

cosA=✓7/4 tanA=3/✓7

I don't know 🙏 🌹 sorry

cosA)=2.645/4 & tanA=3/2.645

sinA- 3/4cosA- 5/4tanA- 3/4

sinA=3/4cosA=5/4tana=3/4

tan A =3/4........................

sinA=√7/4 tanA=3/√7

sinA=3/4CosA-5/4tanA-3/5

cos A=3/√7 and tan A =3\√7

sinA=3/4casA=5/4TanA=3/5

cosA = √7/4tanA = 3/√7

SinA 3/4CosA 5/4TanA 3/5ha Bhai

cosA=√7/4 &tanA=3/√7 sinA=3/4

sin=3/4 L=3, k=4b=√(3^2+4^2) b=5cosA=A/K=5/4Tan=L/A=3/5

sinA=√7/4

tanA=3/√7

CosA√3/4 TanA3/√7 he is my answers

sinA=3/4 cosA=5/4 tanA=3/5

sin A=3/4cos A=5/4tan A=3/4

Sin A =3/4 Cos A =1/4 Tan A =3

sinA=.3/4 cosA=5/4. tanA=3/4

SinA=3/4CosA=root 7/4TanA=3/root 7

sinA= 3/4...so P-7 , B-3 ,H-4

So cosA=7/4& tanA=3/7

value of cosA=4/5 and tanA=3/4

tan A = 3/√7 and cosA= √7/4

sin A =3/4then base=√7then cos A= √7/4 and tan A= 3/√7

this is wrong question because root value of not applicable on p/q form

ANSWERsinA=3/4cosA=5/4tanA =3/4

cos=root7/4tan=3/root7sin=3/4

cosA= √7/4tanA=3/√7

sinA=3/4CosA=√7/3TanA=3/√7

cosA=√7/4TanA=3/√7sona=3/4

cos = 7/4. Tan= 3/7

sinA=3/4 cosA=5/4 tanA=3/4

CosA=root5/4 tanA=3/root5

Sin A= 3/4 = p/h in pithagorius form ( here p=3 & h= 4)So now b = whole root over of h square - P square = root over of 16- 9 = root 7So cos A = b/h = root7/4 &. Tan A = p/b = 3/root7

sinA=3/4 we know sinA=p/hand h2=p2+b2hereh=4andp=3so,16=9+bb=16_9b=7squrethen cos=root7/4and tanA=3/root7

sinA=3/4CosA=5/4tanA=3/4

we know that sinA =3/4,cosA=√7/4sinA/cosA=3/7

cosA= √7/4 , tanA= 3/√7

cosA=√7/4,and tanA=3/√7

cos A= 5/4tan A= 3/5

sinA=3/4cos=5/4tan=3/4

Diya hai sinA= 3/4 jante Hain formula=Lal beti ka ka

given that sinA=3/4 Cos A =√7/4 and tanA=3/√7

cosA=√7÷4and tanA=3÷√7

cos A=5/4 and tanA =3/5

cosA= root7/4 and tanA=3/root7

sin A=3/4 and cosA=√7/4 and tanA=3/√7

cosA= √7/4

tanA= 3/√7