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halosh92:
1)triangle ABC is such that AB=5cm , AC=10 cm and triangle ABC=90 degrees. (angle B=90)
an arc of a circle , center A and radius 5cm cuts AC at D
a) state , in radians the value of angle A in BAC
b)  calculate the are of the region enclosed by BC , DC , and the arc BD


2) a minor sector OMN of a circle center O and radius "r". the perimeter of the sector is 100cm and the area of the sector is A cm^2
a) show that A=50r-r^2
b) given that r varies , find:
i) the value of "r" for which A is a maximum and show that A is a maximum
ii) the value of MON for this maximum area
iii) the maximum area of sector OMN


plzzzz help!

astarmathsandphysics:
2 hours

halosh92:
3) sector OAB of a circle, center O and radisu "r" cm. the length of the arc AB is "p" cm and AOB is "tita" radians..
a) find "tita" in terms of "p" and 'r"
b) deduce that the area of the sector is 1/2pr
given that r= 4.7 and p=5.3, where each has been measured to 1 decimal place, find giving your answer to 3 decimal places.
c) the least possible value of the area of the sector
d) the range of possible values of "tita"

i dono how to do part c and d

astarmathsandphysics:
1)triangle ABC is such that AB=5cm , AC=10 cm and triangle ABC=90 degrees. (angle B=90)
an arc of a circle , center A and radius 5cm cuts AC at D
a) state , in radians the value of angle A in BAC
b)  calculate the are of the region enclosed by BC , DC , and the arc BD
a)pi/3
b)1/2(5*sqrt(75)-5^2)


2) a minor sector OMN of a circle center O and radius "r". the perimeter of the sector is 100cm and the area of the sector is A cm^2
a) show that A=50r-r^2
b) given that r varies , find:
i) the value of "r" for which A is a maximum and show that A is a maximum
ii) the value of MON for this maximum area
iii) the maximum area of sector OMN

a)100=2r+r theta
theta=(100-2r)/r
A=1/2r^2 theta=1/2*r^2*(100-2r)/r=50r-r^2
b)dA/dr=50-2r=0 so r=25
d^2A/dr" =-2<0 hence max
theta=MON=(100-2r)/r=2 when r=25
Area=50r-r^2 =625 when r=25
 

astarmathsandphysics:
c)find 1/max(2pr)=1/(2*4.65*6.35)
d)tita=r/r
find (max p)/(min r) to (min p)/(max r)
ie 5.35/4.65 to 5.25/4.75

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