+0

# Let and be acute angles whose sum is also acute, such that and .Let be the foot of the perpendicular

+2
258
6
+63

$$F$$Let $$\alpha$$ and $$\beta$$ be acute angles whose sum is also acute, such that $$\angle BAC = \alpha, \angle DAB = \beta$$ and $$AB = 1$$.Let $$E$$ be the foot of the perpendicular from $$D$$ to $$AC$$, and  be the foot of the perpendicular from $$B$$ to $$DE$$, as shown below:

a) Prove that $$\triangle DBF$$ is similar to $$\triangle ABC$$.

b) Calculate $$AC, BC, AD, BD, BF$$ and $$DF$$ in terms of trigonometric functions of  $$\alpha$$and $$\beta$$.

c) Use the diagram above to provide another proof of the sine and cosine angle sum identities for acute angles  $$\alpha$$and $$\beta$$whose sum is below$$90^{\circ}$$ .

Jul 25, 2021

#1
+63
+2

Sorry, I meant that F be the foot of the perpendicular, i had no idea how the f went to the front

Jul 25, 2021
#2
+121085
+2

I can do a few

(a)  Angles   BCA   and  DFB     = 90

Angle DBA   = angle  FBC  = 90

Angle DBA   =Angle  DBF  +  Angle FBA

Angle FBC  =  Angle ABC  +  Angle  FBA

Which implies  that angles  DBF  and ABC  are equal

Thus.....by AA congruence.....triangle ABC  is similar to triangle DBF

Jul 25, 2021
#3
+121085
+2

(b)

cos  a   =  AC  /1    ⇒      AC   =cos   a

sin a  = BC  /  1    ⇒   BC   =   sin a

cos b   =  1 / AD   ⇒   AD  =  1/ cos b  ⇒   AD  =  sec  b

tan b =   BD  / 1     ⇒     BD  = tan  b

Not sure  about  the last  two.......but here's  what I  get

angle  a  =   angle  FDB .......so

sin  a   =   sin  FDB

sin a   =    BF  / DB

sin a  =  BF  /  tanb    ⇒     sin a tan b   =  BF

Similarly

cos  a  =  cos   FDB

cos  a   =   DF /  DB

cos  a  =  DF / tan  b   ⇒   cos a  tan b   =    DF

Jul 25, 2021
#5
+63
+2

TYSM!

I was just trying to figure them out, but i couldnt.

Thanks! :)

elloooo  Jul 25, 2021
#6
+121085
+2

Ur  Welcome   !!!!!

CPhill  Jul 25, 2021
#4
+121085
+2

For  the  last one......see if this helps  :    https://themathpage.com/aTrig/sum-proof.htm

Jul 25, 2021