+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\(90^{\circ}\) .
+63 Sorry, I meant that F be the foot of the perpendicular, i had no idea how the f went to the front
+128472 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
+128472 (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
