Question

if 5sec∅-12cosec∅=0,find the values of sec∅,cos∅,and sin∅​

Answer 1

Answer:

$5 \sec(Ø) - 12 \cosec(Ø) = 0 \\ 5 \times \frac{1}{ \cos(Ø) } - 12 \times \frac{1}{ \sin(Ø) } = 0 \\ \frac{5}{ \cos(Ø) } - \frac{12}{ \sin(Ø) } = 0 \\ \frac{5 \sin(Ø) - 12 \cos(Ø) }{ \sin(Ø) \cos(Ø) } = 0 \\ 5 \sin(Ø) - 12 \cos(Ø) = 0 \\ 5 \sin(Ø) = 12 \cos(Ø) \\ \frac{ \ \sin (Ø) }{ \cos(Ø) } = \frac{12}{5} \\ \tan(Ø) = \frac{12}{5}$

TanØ = Opposite side/Adjacent side = 12/5

In a Right-angled triangle,

by pythogoras theorem,

(Hypotenuse)^2 = (Altitude)^2 + (Base)^2

In a ∆ABC right angled at B,

and angle ACB = Ø,

AC^2 = AB^2 + BC^2

AC^2 = 12^2 + 5^2

AC^2 = 144 + 25

AC^2 = 169

AC = 13

Hypotenuse = 13

Thus, if TanØ = 12/5, the values of,

• SecØ = 13/5
• CosØ = 5/13
• SinØ = 12/13