Question 1090326
What did you mean by "to the nearest 10l" ?
Was it a typo and you meant to the nearest {{{10^o}}} ?
Are you allowed to use a calculator?
I believe you are asking for solutions to {{{sec(q)=-1.116}}}
in the interval {{{"["}}}{{{0^o}}}{{{","}}}{{{360^o}}} {{{")"}}} .
The interval {{{"["}}}{{{0^o}}}{{{","}}}{{{360^o}}} {{{")"}}} is the values between {{{0^o}}} (included) and {{{360^o}}} (not included),
so you are looking for solutions with {{{0^o<=q<360^o}}} .
The trigonometric function is secant, which is the reciprocal of cosine, meaning
{{{sec(q)=1/cos(q)}}} .
The graph of the function {{{cos(x)}}} ,
with {{{x}}} measured in degrees looks like the red curve below:
{{{graph(650,300,-50,500,-2,2,cos(x*pi/180),-1,1)}}} .
The graph of the function {{{sec(x)}}} ,
with {{{x}}} measured in degrees looks like the red curve below:
{{{graph(650,300,-50,500,-2,2,1/cos(x*pi/180),-1,1)}}} .
The solutions to {{{sec(q)=-1.116}}}
in the interval {{{"["}}}{{{0^o}}}{{{","}}}{{{360^o}}} {{{")"}}}
are the {{{x}}} values for the two circled points marked below
{{{drawing(650,300,-50,500,-2,2,
line(-50,-1.116,500,-1.116),
locate(300,-1.12,y=-1.116),
circle(153.6,-1.116,5),
circle(206.4,-1.116,5),
graph(650,300,-50,500,-2,2,1/cos(x*pi/180))
)}}} .
Those two points are the only points on the graph with {{{sec(x)=-1,116}}}
and {{{x}}} between {{{0^o}}} and {{{360^o}}} .
Finding solutions:
{{{sec(q)=-1.116}}}
{{{1/cos(q)=-1.116}}}
{{{cos(q)=-1/1.116}}}
{{{cos(q)=approximately}}}{{{-0.8960573}}} .
Using the function inverse cosine in a calculator you can find
the solution for {{{q}}} that is between {{{0^o}}} and {{{180^o}}} ,
(the second quadrant solution):
{{{q=approximately}}}{{{153.6446^o}}} .
So, {{{q=approximately}}}{{{-153.6446^0}}} is third quadrant solution,
but if we want one between {{{180^o}}} and {{{360^o}}} ,
we need to add {{{360^o}}} , to get
{{{q=approximately}}}{{{180^o-153.6446^o=206.3554^o}}} .