Asked by subhankar
If sinA=12/13 & tan B=3/4 then find
¡= sin(A+B)
¡¡=COS(A-B)
¡¡¡=COT(A-B)
¡v=cot(A-B)
Sir please give me the correct ans because tomorrow is my class
¡= sin(A+B)
¡¡=COS(A-B)
¡¡¡=COT(A-B)
¡v=cot(A-B)
Sir please give me the correct ans because tomorrow is my class
Answers
Answered by
Reiny
for sinA = 12/13, we have the 5-12-13 triangle,
so cosA = 5/13
and tanA= 12/5
for tanB = 3/4, we have the 3-4-5 triangle,
so sinB = 3/5
and cosB = 4/5
you have to know the expansion formulas, I will do the first one, you do the rest
sin(A+B) = sinAcosB + cosAsinB
= (12/13)(4/5) + (5/13)(3/5)
= 48/65 + 15/65 = 63/65
always check with a calculator, my answer is correct
I noticed your last two questions are the same, watch for a typo
so cosA = 5/13
and tanA= 12/5
for tanB = 3/4, we have the 3-4-5 triangle,
so sinB = 3/5
and cosB = 4/5
you have to know the expansion formulas, I will do the first one, you do the rest
sin(A+B) = sinAcosB + cosAsinB
= (12/13)(4/5) + (5/13)(3/5)
= 48/65 + 15/65 = 63/65
always check with a calculator, my answer is correct
I noticed your last two questions are the same, watch for a typo
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