Hi, just wondering if anyone can solve this maths problem.
If sin x° = sin 60° and 90°< x <180°, find x.
Thanks.
Please show working; very appreciated.
as sin x = sin (180- x)
so if 180-x= 60 then x = 120 so 120 is the answer
60 X 180
for sine function, the value of x is always the same for 180-x since the function is symmetrical with respect to y-axis..
therefore sin(x)=sin(180-x)
if x=60 degrees, then
180-x=120 degrees..
which satisfies the condition 90 degrees<x<180 degrees
sin(60) = sin(180 - 60) = sin(120)
x = 120
120
remember the circle
120+60=180
so both of the sin's of these degree's are rad3/2
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as sin x = sin (180- x)
so if 180-x= 60 then x = 120 so 120 is the answer
60 X 180
for sine function, the value of x is always the same for 180-x since the function is symmetrical with respect to y-axis..
therefore sin(x)=sin(180-x)
if x=60 degrees, then
180-x=120 degrees..
which satisfies the condition 90 degrees<x<180 degrees
sin(60) = sin(180 - 60) = sin(120)
x = 120
120
remember the circle
120+60=180
so both of the sin's of these degree's are rad3/2