THe interior angles of an n-sided convex polygon are in AP. If the common difference is 5 and the largest angle is 160, then n equals? @ganeshie8
lets try and setup two equations in "n" and "a"
largest angle is 160 :\[a+(n-1)5 = 160\tag{1}\]
interior angles add up to 180(n-2) : \[\dfrac{n}{2}\left[2a+(n-1)5\right] = 180(n-2)\tag{2}\]
two equations and two unknowns, you can solve them
one sec
its a bit trickier than the previous problem as you can see the second equation is quadratic, so you will need to do a bit more work
cant get it
show me
try to substitute the value of "a" from first equation into second equation
i tried but it doesnt seem to work
wait
ok waiting
is it 16
@ganeshie8
doesn't look correct
isolating "a" from first eqn gives you \[a = 160-5(n-1)\] substitute that in second eqn
\[\dfrac{n}{2}\left[2(160-5(n-1))+(n-1)5\right] = 180(n-2)\] simplify a bit and you will see a nice quadratic in "n"
what is it
i cant seem to get it
have you understood everythign so far ?
yes
Okay do you know to use distributive property and combine like terms ?
yes
i just cant gettit for some strange reason
then whats stopping you from simplifying the above mess ?
\[\dfrac{n}{2}\left[2(160-5(n-1))+(n-1)5\right] = 180(n-2)\] start by distrubuting \[\dfrac{n}{2}\left[2\times 160-2\times 5(n-1))+5(n-1)\right] = 180\times n-180\times 2\]
i know i didnt get a quadratic but got n=18 which is wrong
\[\dfrac{n}{2}\left[2\times 160-5(n-1)\right] = 180n-360\]
\[\dfrac{n}{2}\left[325-5n\right] = 180n-360\]
multiply 2 both sides \[n\left[325-5n\right] = 360n-720\]
the roots are -16 and 9
so its 9 oooops forgot to redo the minus
Yep! discard -16 as the number of sides cannot be negative
The key part is setting up the equations correctly using the properties of AP and triangles. Once you have the equations, you can solve them using any of your favorite methods
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