SOLUTION: The 45degree 60degree 75degree triangle has angles in arithmetic sequence. In triangle ABC, AB=1, angle A=45degrees, and angle B=60 degrees, without using a calculator how can I s

Algebra ->  Trigonometry-basics -> SOLUTION: The 45degree 60degree 75degree triangle has angles in arithmetic sequence. In triangle ABC, AB=1, angle A=45degrees, and angle B=60 degrees, without using a calculator how can I s      Log On


   

Question 167514This question is from textbook Functions and Relations
: The 45degree 60degree 75degree triangle has angles in arithmetic sequence.
In triangle ABC, AB=1, angle A=45degrees, and angle B=60 degrees, without using a calculator how can I show these expressions?
a)the lengths of AC and BC
b)sin 75degrees, cos 75degrees, tan75degrees
c)sin 15degrees, cos 15degrees, tan 15degrees This question is from textbook Functions and Relations

Answer by jojo14344(1513) About Me  (Show Source):
You can put this solution on YOUR website!
I'll leave you the sketch, hopefully we can follow together.
In the triangle ABC, given:
A=45%5Eo; B=60%5Eo; C=75%5Eo
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.
---> BLUE circle>>>A=45%5Eo; GREEN Circle>>> B=60%5Eo; RED Circle>>> C=75%5Eo. 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:
---> BLACK Line=BD
We know AC=AD%2BDC -------------> Eqn 1 >>>> as shown on the graph
We find ABD forms right triangle. By Trigo Functions:
Find first Line AD ---> tan45%5Eo=opp%2Fadj=AD%2FBD--->AD=tan45%5Eo%28BD%29---------> Eqn 2:BD=unknown
But, sin45%5Eo=BD%2FAB --> BD=sin45%5Eo%28AB%29--------------> Eqn 3
Subst. Eqn 3 in Eqn 2:
AD=%28tan45%5Eo%29%28sin45%5Eo%29%28AB%29 ------------------> Eqn 4
.
Next, Line DC, On Angle C=75%5Eo:
tan75%5Eo=opp%2Fadj=BD%2FDC
DC=BD%2F%28tan75%5Eo%29=Eqn3%2F%28tan75%5Eo%29=sin45%5Eo%28AB%29%2F%28tan75%5Eo%29 -----------> Eqn 5
Therefore, Via Eqn 1:
------------------> Eqn 6
*Note: AB=1 as per given
.
Next, Line BC=?
cos75%5Eo=adj%2Fhyp=DC%2FBC
BC=DC%2Fcos75%5Eo=Eqn.5%2Fcos75%5Eo
BC=%28%28sin45%5Eo%29%28AB%29%2Ftan75%5Eo%29%29%2Fcos75%5Eo
highlight%28BC=%28sin45%5Eo%29%28AB%29%2F%28tan75%5Eo%29%28cos75%5Eo%29%29 ---------------> Eqn 7
.
b)sin 75degrees, cos 75degrees, tan75degrees
First sin75%5Eo=?,
On Right Triangle CBD with C=75%5Eo:
sin75%5Eo=opp%2Fhyp=BD%2FBC=Eqn.3%2FEqn.7


highlight%28sin75%5Eo=%28tan75%5Eo%29%28cos75%5Eo%29%29
Next, cos75%5Eo=?,
cos75%5Eo=adj%2Fhyp=DC%2FBC=Eqn.5%2FEqn.7


highlight%28cos75%5Eo=cos75%5Eo%29
.
Next, tan75%5Eo=?
tan75%5Eo=opp%2Fadj=BD%2FDC=Eqn.3%2FEqn.5
tan75%5Eo=%28%28sin45%5Eo%29%28AB%29%29%2F%28%28sin45%5Eo%29%28AB%29%2Ftan75%5Eo%29

highlight%28tan75%5E0=tan75%5Eo%29
.
c)sin 15degrees, cos 15degrees, tan 15degrees
We draw a line from A to BD as shown:
Mark Line AE.See below:
----> We call it angle A%5B1%5D=15%5Eo from A --> E.
.
We solve for AE:
highlight%28cos15%5Eo=adj%2Fhyp=AD%2FAE%29
highlight%28AE=AD%2Fcos15%5Eo=Eqn.4%2Fcos15%5Eo%29 ------------> Eqn 8
Next, sin15%5Eo=?
highlight%28sin15%5Eo=opp%2Fhyp=ED%2FAE=ED%2FEqn.8%29
highlight%28ED=%28sin15%5Eo%29%28Eqn.8%29%29 -------------------> Eqn 9
Next, tan15%5Eo=?
highlight%28tan15%5Eo=opp%2Fadj=AD%2FED=Eqn.4%2FEqn.9%29
Try to finish it off with what we've computed on the above EQUATIONS 1-9:
.
Thank you,
Jojo