Mathematics Question and Solution

By Ogheneovo Daniel Ephivbotor
6th June, 2025

Let a be the blue length.


b² = 4²+3²

b = √(25)

b = 5 units.


Let alpha be c.


cos(2c) = 4/5

cos(c+c) =4/5

(cosc)²-(sinc)² = 4/5 --- (1).


cosc = 3/a --- (2).


a² = 3²+d²

d = √(a²-9) units.


sinc = d/a

sinc = √(a²-9)/a --- (3).


Substituting (2) and (3) in (1).


(3/a)²-(√(a²-9)/a)² = 4/5


(9/a²)-((a²-9)/a²) = 4/5


9-(a²-9) = ⅕(4a²)


18-a² = ⅕(4a²)


90-5a² = 4a²


9a² = 90


a² = 10


a = √(10) units.

a = 3.16227766017 units.

Again, a is the required blue length.


Or


tan(2c) = (3/4)

2c = atan(3/4)

c = ½(atan(3/4))

c = 18.4349488229°

c is alpha.


It implies, the required length, a is;


cos(18.4349488229) = 3/a

a = 3/cos(18.4349488229)

a = 3.16227766017 units.

a = √(10) units.

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