Given that 0≤β<2π, find all the solutions to the equation 2sin^2β=7sinβ+4.

Step-by-step explanation:

2sin²β-7sinβ-4=0

2sin²β-8sin β+1 sin β-4=0

2 sin β(sin β-4)+1(sin β-4)=0

(sin β-4)(2sin β+1)=0

either sin β-4=0

sin β=4 (rejected,as |sin β|≤1)

or

2 sin β+1=0

sin β=-1/2=-sin (π/6)=sin (π+π/6),sin (2π-π/6)

β=7π/6,11π/6

d. bc = 12 and ef = 16

step-by-step explanation:

we are given that δabc : δdef such that .

also, it is given that

so, from the options, we need that .

a.

b.

c.  

d.

as, we can see that only option d gives the same ratio as i.e. .

hence, the possible lengths are bc = 12 and ef = 16.