In ΔABC, ∠C is a right angle. If cosB = 5/13, what is sinA?

Answer:
The value of sinA is [tex]\frac{5}{13}[/tex].
Step-by-step explanation:
ΔABC
Base = b = BC
Hypotenuse = h = AB
Perpendicular = p = AC
[tex]\cos B=\frac{b}{h}=\frac{BC}{AB}=\frac{5}{13}[/tex]
If we rotate the triangle in such a way so that base changes from BC to AC, then:
Base = b = AC
Hypotenuse = h = AB
Perpendicular = p = BC
[tex]\sin A=\frac{p}{h}=\frac{BC}{AB}=\frac{5}{13}[/tex]
The value of sinA is [tex]\frac{5}{13}[/tex].
Answer:
The correct answer is Sin A =5/13
Step-by-step explanation:
Trigonometric ratio:-
Sin∅ = opposite side /Hypotenuse
cos∅ = Adjacent side/ Hypotenuse
From the given figure, we get
Triangle ABC is a right angled triangle,
<C = 90°
To find sinA
Cos B = Adjacent side/ Hypotenuse = 5/13
Adjacent side = 5 and Hypotenuse = 13
Adjacent side of angle B = Opposite side of angle A
Therefore,
Sin A = opposite side /Hypotenuse = 5/13
The correct answer is Sin A =5/13