Question

plzz answer it as fast as possible​

Answer 1

Step-by-step explanation:

This implies that

x2+2ax=4x−4a−13

or

x2+2ax−4x+4a+13=0

or

x2+(2a−4)x+(4a+13)=0

Since the equation has just one solution instead of the usual two distinct solutions, then the two solutions must be same i.e. discriminant = 0.

Hence we get that

(2a−4)2=4⋅1⋅(4a+13)

or

4a2−16a+16=16a+52

or

4a2−32a−36=0

or

a2−8a−9=0

or

(a−9)(a+1)=0

So the values of a are −1 and 9.

Answer 2

Formula to be used,

a/sin A=b/sin B=c/sin C, A+B+C =180, sin (180-A) = sin A,

sin (90-A) = cos A, cos (90-A) = sin A,

sin A+sin B = sin (A+B)/2*cos(A-B)/2, sin A-sin B = sin (A-B)/2*cos(A+B)/2

(a-b)/c =(sin A-sin B)/sin C =(sin A-sin B)/sin(A+B) as C =180-(A+B)

(a-b)/c =2sin [(A-B)/2]*cos[(A+B)/2]/2/sin [(A+B)/2]*cos[(A+B)/2]

=sin [(A-B)/2]/sin [(A+B)/2]=sin [(A-B)/2]/sin (90-C/2)=sin [(A-B)/2]/cos(C/2)

Similarly,

a/(b+c)=sin A/(sin B+sin C)=sin (B+C)/(sin B+sin C)=

2sin [(C+B)/2]*cos[(C+B)/2]/2/sin[(B+C)/2]/cos[(B-C)/2]

=cos[(C+B)/2]/cos[(B-C)/2]=sin (A/2)/cos[(B-C)/2]