Homework Statement
##\sin a + \cos b## = ##\frac{-1}{2}##
##\cos a + \sin b## = ##\frac{\sqrt 3}{2}##
0 < a < ##\pi/2##
##\pi/2## < b < ##\pi##
a + b = ? By calculating sin (a+b)
Homework Equations
The Attempt at a Solution
I tried :
##\sin a + \cos b =...
When i examined it carefully. When i realized you substituted the ##\vec a ##. ##\vec {2b}## of #15 into |##\vec a##| |##\vec 2b##|cos##\theta## and resulted the formula in #11. It makes senses...
In vector addition for the law of cosines, we define c with a and b. So, it supposed to be ##\vec c = \vec a + \vec b## right?
a+ b is head of vec a meets tail of vec b right? Or, when
you said that ##\vec c = \vec a - \vec b##. Do you mean that is when head of vec a meets the tail of vec b?
Because i thought finding angle between ##\vec a## and ##\vec { -2b}## i can find angle between a and b. Since 180 - angle(a-2b) = angle(a+b) under the condition, if it's true that ##\vec b## has same degree with regards to x-axis as ##\vec {-2b}## just different direction going left or right...
But it still makes more sense to me. Because the |a| |b| and |a-2b| can be seen as triangle right? |a| is a lenght of vector a with magnitude 2 right?
Although, i find that expanding (a-2b)^2 also can be the way. Maybe if being elaborated, you mean that :
(a-2b)•(a-2b) = |a-2b|.|a-2b|cos0
Thus...
Homework Statement
|a| = 2
|b| = ## \sqrt3##
|a - 2b| = 2
Angle between a and b
Homework Equations
The Attempt at a Solution
##\theta## is angle between a and b
So angle between a and -2b is 180-##\theta## [/B]
##|a-2b|^2## = |a|^2 + |2b|^2 -2|a||2b|cos(180-##\theta##)
##2^2## = 2^2 +...
Hm yes. Then, i am confused how to get 0 < a<= 3 itself..
How i get it is when x=-2, a must be 4
When x = 5, a must be 3
When x = 2, a must be 0
The range for a must be between 0 and 4
Sorry again.. the value of x for x^2 -3x -10<0 is -2<x<5
When -2<x<2
a value is 0<a<4
When 2<=x<5
0<=a<3
But i cannot decide what is the sufficient and necessary conditions
My answer now, sufficient 0<=a<3
Homework Statement
|x-2| < a is a necessary condition for x^2 -3x -10 < 10 . What is the range value of a?
|x-2| < a is a sufficient condition for x^2 -3x -10 < 10. What is the range value of a?[/B]
The options are
a>= 4
a>=3
0<a<=2
0<a<=3
0<a<=5
The Attempt at a Solution
Range of x for x^2...