Q5 c)i)1) (1 Viewer)

[Thecorey0]

will i get it rong if i put theta was equal to 7pie/3
when subbedin it still gives an area of root 3?

Yes. This is when you have to think logically. 7pi/3 > 360 degrees > a whole circle. Hence it is impossible.

 

[oly1991]

that "value for theta" tripped me out so hard....it seemed like an easy 2 marks to get but i didnt know what the other value was, since everything i was working out produced the same result i.e. pie/3

so i went to the next value of cos@ = 1/2, and it was 5pie/3 (300 degrees) *sigh*

to the x.addayeee dude, how the hell did u get 2pie/3??

 

[Thecorey0]

that "value for theta" tripped me out so hard....it seemed like an easy 2 marks to get but i didnt know what the other value was, since everything i was working out produced the same result i.e. pie/3

so i went to the next value of cos@ = 1/2, and it was 5pie/3 (300 degrees) *sigh*

to the x.addayeee dude, how the hell did u get 2pie/3??

Area of triangle = 1/2*absinC = 2sin@
Therefore root(3)/2 = sin@
@ = 2pi/3

 

[oly1991]

Area of triangle = 1/2*absinC = 2sin@
Therefore root(3)/2 = sin@
@ = 2pi/3

hmm
but when sin@ = (root 3)/2

doesnt @ = pi/3 haha

unless my calculator is fucked up

 

[Thecorey0]

hmm
but when sin@ = (root 3)/2

doesnt @ = pi/3 haha

unless my calculator is fucked up

Yes, but the question specifies that there are 2 solutions, hence you take the next solution. Therfore @ = pi - pi/3 = 2pi/3

 

[MetroMattums]

Or maybe. Given the angle is pi/3. And you know the radii. It's an equilateral triangle.

ie. The chord length is 2. Don't see why the cosine rule is necessary..