PLS I NEED HELP ASAP ILL GIVE 20 POINTS
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It has not been indicated whether the figure in the questions is a triangle or a quadrilateral. Irrespective of the shape, this can be solved. The two possible shapes and angles have been indicated in the attached image.
Now, from the information given we can infer that there is a line BD that cuts angle ABC in two parts: angle ABD and angle DBC
⇒ Angle ABC = Angle ABD + Angle DBC
Also, we know that angle ABC is 1 degree less than 3 times the angle ABD, and that angle DBC is 47 degree
Let angle ABD be x
⇒ Angle ABC = 3x-1
Also, Angle ABC = Angle ABD + Angle DBC
Substituting the values in the above equations
⇒ 3x-1 = x+47
⇒ 2x = 48
⇒ x = 24
So angle ABD = 24 degree, and angle ABC = 3(24)-1 = 71-1 = 71 degree
Take derivitive
note
the derivitive of sec(x)=sec(x)tan(x)
so
remember the quotient rule
the derivitive of
so
the derivitive of
so now evaluate when t=pi
we get
sec(pi)=-1
tan(pi)=0
we get
slope=1/pi
use slope point form
for
slope=m and point is (x1,y1)
equation is
y-y1=m(x-x1)
slope is 1/pi
point is (pi,1/pi)
y-1/π=1/π(x-π)
times both sides by π
πy-1=x-π
πy=x-π+1
y=(1/π)x-1+(1/π)
or, alternately
-(1/π)x+y=(1/π)-1
x-πy=π-1
note
the derivitive of sec(x)=sec(x)tan(x)
so
remember the quotient rule
the derivitive of
so
the derivitive of
so now evaluate when t=pi
we get
sec(pi)=-1
tan(pi)=0
we get
slope=1/pi
use slope point form
for
slope=m and point is (x1,y1)
equation is
y-y1=m(x-x1)
slope is 1/pi
point is (pi,1/pi)
y-1/π=1/π(x-π)
times both sides by π
πy-1=x-π
πy=x-π+1
y=(1/π)x-1+(1/π)
or, alternately
-(1/π)x+y=(1/π)-1
x-πy=π-1