in ]-π , 2π[

#### Subhotosh Khan

##### Super Moderator
Staff member
Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

#### nessmell

##### New member
Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

i m sorry i m new here ... i don't know where to start

#### Subhotosh Khan

##### Super Moderator
Staff member
i m sorry i m new here ... i don't know were to start
Have you studied "unit circle"?

#### HallsofIvy

##### Elite Member
Can you do the problem now? I presume you know that an "equilateral triangle" has a all sides the same length (let's call th\fat "s") and all angles the same measure (so 180/3= 60 degrees). The line from one vertex perpendicular to the opposite side bisects that side and bisects the angle- so it divides the equilateral triangle into two 30- 60- 90 right triangles. The hypotenuse of each of those right triangles is s while the side opposite the 30 degree angle is s/2. By the Pythagorean theorem, the side opposite the 60 degree triangle has length t that satisfies $$\displaystyle t^2+ (side/2)^2= s^2$$ so $$\displaystyle t^2= s^2- (s/2)^2= s^2- s^2/4= s^2(1- 1/4)= s^2(3/4)$$ so that $$\displaystyle t= \frac{\sqrt{3}}{2}s$$. The side opposite the 60 degree angle (so next to the 30 degree angle) divided by the hypotenuse is $$\displaystyle \frac{\sqrt{3}}{2}$$.

#### nessmell

##### New member
Can you do the problem now? I presume you know that an "equilateral triangle" has a all sides the same length (let's call th\fat "s") and all angles the same measure (so 180/3= 60 degrees). The line from one vertex perpendicular to the opposite side bisects that side and bisects the angle- so it divides the equilateral triangle into two 30- 60- 90 right triangles. The hypotenuse of each of those right triangles is s while the side opposite the 30 degree angle is s/2. By the Pythagorean theorem, the side opposite the 60 degree triangle has length t that satisfies $$\displaystyle t^2+ (side/2)^2= s^2$$ so $$\displaystyle t^2= s^2- (s/2)^2= s^2- s^2/4= s^2(1- 1/4)= s^2(3/4)$$ so that $$\displaystyle t= \frac{\sqrt{3}}{2}s$$. The side opposite the 60 degree angle (so next to the 30 degree angle) divided by the hypotenuse is $$\displaystyle \frac{\sqrt{3}}{2}$$.
thank you ... we have -√3/2 not √3/2

#### Subhotosh Khan

##### Super Moderator
Staff member
In which quadrant (I or II or III or IV), the value of "cosine" is negative (< 0)?

#### nessmell

##### New member
In which quadrant (I or II or III or IV), the value of "cosine" is negative (< 0)?

#### HallsofIvy

##### Elite Member
Cute! So now you know, from post, from post #9. that cos(30)=√32cos(30)=32. Now you know that cosine is negative in the second and third quadrants. cos(180- 30)= -cos(30)