Step-by-step explanation:
Given that,
Hypotenuse, H = 13
Base, b = 12
Perpendicular, P = 5
[tex]\sin\theta=\dfrac{P}{H}\\\\\sin\theta=\dfrac{5}{13}[/tex]
[tex]\cos\theta=\dfrac{B}{H}\\\\\sin\theta=\dfrac{12}{13}[/tex]
and
[tex]\tan\theta=\dfrac{P}{B}\\\\=\dfrac{5}{12}[/tex]
Hence, this is the required solution.
