Question 167514
I'll leave you the sketch, hopefully we can follow together.
In the triangle ABC, given:
{{{A=45^o}}}; {{{B=60^o}}}; {{{C=75^o}}}
{{{AB=1unit}}}
Pardon me for the illustration below is just for reference --> NOT TO SCALE. So it will be easy for us to go along.
{{{drawing(300,300,-5,15,-5,15,graph(300,300,-5,15,-5,15),blue(line(1,1,12,1)),locate(2,2.3,A),green(line(1,1,9,12)),locate(8.5,10.5,B),red(line(12,1,9,12)),locate(11,2.3,C),blue(circle(1,1,.30)),green(circle(9,12,.30)),red(circle(12,1,.30)))}}} ---> BLUE circle>>>{{{A=45^o}}}; GREEN Circle>>> {{{B=60^o}}}; RED Circle>>> {{{C=75^o}}}. Also,BLUE LINE ={{{AC}}}; GREEN LINE={{{AB}}}; RED LINE={{{BC}}}

a)the lengths of AC and BC
In order to find {{{AC}}} and {{{BC}}} with given data,w e draw a line from {{{B}}} to line {{{AC}}} and mark LINE {{{BD}}}. See below:
{{{drawing(300,300,-5,15,-5,15,graph(300,300,-5,15,-5,15),blue(line(1,1,12,1)),locate(2,2.3,A),green(line(1,1,9,12)),locate(8.5,10.5,B),red(line(12,1,9,12)),locate(11,2.3,C),blue(circle(1,1,.30)),green(circle(9,12,.30)),red(circle(12,1,.30)),line(9,1,9,12),locate(8.3,2.3,D))}}} ---> BLACK Line={{{BD}}}
We know {{{AC=AD+DC}}} -------------> Eqn 1 >>>> as shown on the graph
We find {{{ABD}}} forms right triangle. By Trigo Functions:
Find first Line {{{AD}}} ---> {{{tan45^o=opp/adj=AD/BD}}}--->{{{AD=tan45^o(BD)}}}---------> Eqn 2:{{{BD=unknown}}}
But, {{{sin45^o=BD/AB}}} --> {{{BD=sin45^o(AB)}}}--------------> Eqn 3
Subst. Eqn 3 in Eqn 2:
{{{AD=(tan45^o)(sin45^o)(AB)}}} ------------------> Eqn 4
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Next, Line {{{DC}}}, On Angle {{{C=75^o}}}:
{{{tan75^o=opp/adj=BD/DC}}}
{{{DC=BD/(tan75^o)=Eqn3/(tan75^o)=sin45^o(AB)/(tan75^o)}}} -----------> Eqn 5
Therefore, Via Eqn 1:
{{{highlight(AC=Eqn.4+Eqn.5=(tan45^o)(sin45^o)(AB)+(sin45^o(AB))/(tan75^o))}}}------------------> Eqn 6
*Note: {{{AB=1}}} as per given
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Next, Line {{{BC}}}=?
{{{cos75^o=adj/hyp=DC/BC}}}
{{{BC=DC/cos75^o=Eqn.5/cos75^o}}}
{{{BC=((sin45^o)(AB)/tan75^o))/cos75^o}}}
{{{highlight(BC=(sin45^o)(AB)/(tan75^o)(cos75^o))}}} ---------------> Eqn 7
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b)sin 75degrees, cos 75degrees, tan75degrees 
First {{{sin75^o}}}=?,
On Right Triangle {{{CBD}}} with {{{C=75^o}}}:
{{{sin75^o=opp/hyp=BD/BC=Eqn.3/Eqn.7}}}
{{{sin75^o=(sin45^o(AB))/(((sin45^o(AB))/(tan75^o)(cos75^o)))}}}
{{{sin75^o=(cross(sin45^o(AB))/cross(sin45^o(AB)))((tan75^o)(cos75^o))}}}
{{{highlight(sin75^o=(tan75^o)(cos75^o))}}}
Next, {{{cos75^o}}}=?,
{{{cos75^o=adj/hyp=DC/BC=Eqn.5/Eqn.7}}}
{{{cos75^o=((sin45^o)(AB)/tan75^o)/((sin45^o)(AB)/(tan75^o)(cos75^o))}}}
{{{cos75^o=((cross(sin45^o)cross(AB))/cross(tan75^o))(cross(tan75^o)(cos75^o)/cross(sin45^o)cross(AB))}}}
{{{highlight(cos75^o=cos75^o)}}}
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Next, {{{tan75^o}}}=?
{{{tan75^o=opp/adj=BD/DC=Eqn.3/Eqn.5}}}
{{{tan75^o=((sin45^o)(AB))/((sin45^o)(AB)/tan75^o)}}}
{{{tan75^o=(cross(sin45^o)cross(AB))*(tan75^o)/cross(sin45^o)cross(AB))}}}
{{{highlight(tan75^0=tan75^o)}}}
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c)sin 15degrees, cos 15degrees, tan 15degrees
We draw a line from {{{A}}} to {{{BD}}} as shown:
Mark Line {{{AE}}}.See below:
{{{drawing(300,300,-5,15,-5,15,graph(300,300,-5,15,-5,15),blue(line(1,1,12,1)),locate(2,2.3,A),green(line(1,1,9,12)),locate(8.2,10.5,B),red(line(12,1,9,12)),locate(11,2.3,C),blue(circle(1,1,.30)),green(circle(9,12,.30)),red(circle(12,1,.30)),line(9,1,9,12),locate(8.2,2.3,D),line(1,1,9,3),locate(8.2,3.9,E))}}} ----> We call it angle {{{A[1]=15^o}}} from A --> E.
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We solve for {{{AE}}}:
{{{highlight(cos15^o=adj/hyp=AD/AE)}}}
{{{highlight(AE=AD/cos15^o=Eqn.4/cos15^o)}}} ------------> Eqn 8
Next, {{{sin15^o}}}=?
{{{highlight(sin15^o=opp/hyp=ED/AE=ED/Eqn.8)}}}
{{{highlight(ED=(sin15^o)(Eqn.8))}}} -------------------> Eqn 9
Next, {{{tan15^o}}}=?
{{{highlight(tan15^o=opp/adj=AD/ED=Eqn.4/Eqn.9)}}}
Try to finish it off with what we've computed on the above EQUATIONS 1-9:
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Thank you,
Jojo