Here's the question you clicked on:
eroshea
The axis of the cone 2m long makes an angle of 65 degrees with its horizontal base. What is the area of the base if its volume is 5 cu. m?
|dw:1355654103593:dw| Guessing it looks like this...
but it is not a right triangle so i can't use SOH CAH TOA there
You still can. Notice the height h makes a right angle with the base. |dw:1355654861209:dw| Volume of a cone is 1/3x(area of base)x(height) or \[V = \frac{ 1 }{ 3 } B h\] B is area of the base, h is height. We can find height from the diagram: \[\sin 65 = \frac{ h }{2 }\] so h=2sin65 and we know V = 5 cm^3 \[5 = \frac{ 1 }{ 3 } B \times 2 \sin 65\]
This diagram may help: http://upload.wikimedia.org/wikipedia/commons/d/d2/Cone_3d.png notice the cone on the right, h makes a right angle with the base, and we know the angle that the axis makes with the base.
B is 8.28 and B = pi r^2 root 8.28/pi = r ??? is that correct?
Remember that the question is asking for the area of the base, which is the B. It's not asking for the radius.
yes.. but i'm just asking whether what i did up there is right or wrong..
I got B as 8.275 m^2, so you're correct. Don't forget the units, metres squared. btw i made a mistake up there, i said V = 5 cm^3 but it's m^3 not cm^3.
that's fine.. thanks for helping :))
If you were asked for the radius, then you'd do as you did: \[r = \sqrt{\frac{ 8.275 }{ \pi }}\]