3tanA= 4 then find sinA and cosA
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solution: 3tan=4
then, tan=4/3=. (tan=perpendicular/
hight.)
then, we should find out the base by using Pythagoras theoram.
(h)2=(p)2+(b)2
(3)2=(4)2+(b)2
9=16+(b)2
(b)2= 16-9
(b)2= 7
b=√7
then sin= p\b,cos a =b/h
sina= p/b=4/√7
cosa=b/h=√7/3.
then, tan=4/3=. (tan=perpendicular/
hight.)
then, we should find out the base by using Pythagoras theoram.
(h)2=(p)2+(b)2
(3)2=(4)2+(b)2
9=16+(b)2
(b)2= 16-9
(b)2= 7
b=√7
then sin= p\b,cos a =b/h
sina= p/b=4/√7
cosa=b/h=√7/3.
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