Is a triangle with side lengths 6, 10, 12 a right triangle?
2 answers:
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Quick vocab:
sin=sine
cos=cosine
tan=tangent
theta=the angle
o=opposite side from angle
h=hypotenuse of triangle
a=adjacent side to angle
Three equations:
sin(theta)= o / h
cos(theta)= a / h
tan(theta)= o / a
Explanation of problem:
sin(x)= o / h
sin(x)= 16cm / 28cm
You'll substitute the 16cm in for o and 28cm for h
This is because 16cm is on the side that's opposite from the angle
28cm is on the hypotenuse of the triangle
One way to find the hypotenuse is look where the 90° is at and it's opposite from that
since you're trying to find the angle you'll have to inverse the sin to get it on the other side of the equal sign
Inverse is doing the opposite of so the inverse of sin is sin^-1
x = sin^-1 (16cm/28cm)
Answer:
x = 34.84990458°
I don't know how far rounded you'll need it
I hope this helped!
sin=sine
cos=cosine
tan=tangent
theta=the angle
o=opposite side from angle
h=hypotenuse of triangle
a=adjacent side to angle
Three equations:
sin(theta)= o / h
cos(theta)= a / h
tan(theta)= o / a
Explanation of problem:
sin(x)= o / h
sin(x)= 16cm / 28cm
You'll substitute the 16cm in for o and 28cm for h
This is because 16cm is on the side that's opposite from the angle
28cm is on the hypotenuse of the triangle
One way to find the hypotenuse is look where the 90° is at and it's opposite from that
since you're trying to find the angle you'll have to inverse the sin to get it on the other side of the equal sign
Inverse is doing the opposite of so the inverse of sin is sin^-1
x = sin^-1 (16cm/28cm)
Answer:
x = 34.84990458°
I don't know how far rounded you'll need it
I hope this helped!