1) One leg of a right triangle has length 20 and the radius of the inscribed circle is 6. Compute the length of the hypotenuse of the right triangle. Answer: 29 2) Let a be a positive number...
1) One leg of a right triangle has length 20 and the radius of the inscribed circle is 6. Compute the length of the hypotenuse of the right triangle.
2) Let a be a positive number such that a^2009-2a + 1 = 0: Find all possible values of the sum S = 1 + a + a^2 +… + a^2008:
Answer: 2 & 2009 (I can find 2009 but I can’t find out how to get 2)
3) Suppose (a + bi)^2 = 4(b + ai); where a > 0; b > 0; and i =rad -1; If a + bi =r(cosx + isin x), what is the ordered pair (r,x)? ( x should be in radians)
Answer:( 4, pi/6)
Only one question can be answered either 1 or 2 or 3.
1.the length of tangents from exterior point to circle are equal in length.
So leg of right angle triangle legth 20 can be written as 6+14 ( radius of inscibed cirle is 6). another leg can be written as x+6 and hypotenuse can be wriiten as 14+6 ,where x is length to be determined. By Pythagoras theorem
length of hypotenuse= 14+15=29
2.It is GP ,first term is 1,common ration a and number of terms 2009
SUM= 1.(a^2009-1)/(a-1) ,provided a >1 (i)
substituting a^2009 from (ii) in (i)
compare real and imaginary parts we have
from (i) and (ii) we get
b=2 and a=sqrt(12)
now sqrt(12)+2i= rcos(x)+irsin(x)
comparing real and imaginary parts we have
r cos(x)= sqrt(12) (iii)
r sin(x)= 2 (iv)
sqraring adn adding (iii) and (iv) ,we have
divide (iv) by (iii),we have