some help please

[Calculator]

If the radius of the circle is 10 cm, and the inscribed angle is 55 degrees, find the length of the subtended chord.

This is what I did:
tan(55) = h/5
h = 7, 14 cm

and now using the Pythagorean theorem: x^2 =h^2+5^2
x =8, 7 cm

where h is the height of the triangle,
and x is the chord(i.e. one side of the triangle?)

Is this correct?

 

[Deleted member 4993]

If the radius of the circle is 10 cm, and the inscribed angle is 55 degrees, find the length of the subtended chord.

This is what I did:
tan(55) = h/5
h = 7, 14 cm

and now using the Pythagorean theorem: x^2 =h^2+5^2
x =8, 7 cm

where h is the height of the triangle,
and x is the chord(i.e. one side of the triangle?)

Is this correct?

Subtended angle with center = 2* inscribed angle

sin (55°) = (h/2)/10

 

[Quaid]

the radius of the circle is 10 cm, and the inscribed angle is 55 degrees

find the length of the subtended chord

tan(55) = h/5

h = 7, 14 cm

and now using the Pythagorean theorem: x^2 = h^2 + 5^2

x = 8, 7 cm

where h is the height of the triangle, and x is the chord (i.e. one side of the triangle?)

Is this correct?

Hi Calculator:

No, the triangle with sides 10, 10, and x is not a right triangle. You need right triangles, for Pythagoras and Tangent (as you've used them).

Here's some hints. You're asked for x, not h. Draw a sketch. You've got a triangle with equal sides 10 and base x. If you bisect that triangle, you'll end up with two congruent right-triangles.

Use the sine function, with one of these right triangles, to write the ratio of the side opposite the 55-degree angle (its length is x/2) to the hypotenuse (its length is 10).

Then solve for your x.

Cheers

 

[Calculator]

Hi Calculator:

No, the triangle with sides 10, 10, and x is not a right triangle. You need right triangles, for Pythagoras and Tangent (as you've used them).

Here's some hints. You're not asked for h. Draw a sketch. You've got an isosceles triangle with equal sides 10 and base x. If you bisect that triangle, you'll end up with two congruent right-triangles.

Use the sine function, with one of these right triangles, to write the ratio of the side opposite the 55/2-degree angle (its length is x/2) to the hypotenuse (its length is 10).

Then solve for your x.

Cheers

Thank you for trying to explain it to me, but I didn't understand a thing you said and obviously everything I did is wrong. I don't even know how to draw a sketch

 

[Quaid]

I don't even know how to draw a sketch

I fixed some typos, in my first reply. Sorry for any confusion.




The rays of the inscribed angle are in blue, and the rays of the central angle are in red.

The subtended chord is shown as a thick, grey line segment. You're trying to find the length of the chord. You named this length x.

If we bisect the triangle with red sides (those side lengths are equal to the circle's radius, by the way), then both the 110° angle and the chord are cut in half and we get two identical right triangles. Use one of these right triangles.

Its hypotenuse is exactly the same as the radius of the circle.

The 110° angle got cut in half, so the corresponding angle inside the right triangle is 110°/2, or 55° (same as the blue, inscribed angle -- that always happens).

The length of the right-triangle's side that is opposite the (110°)/2 angle is x/2.

The right-triangle trigonometric ratio for sine is sin(angle)=[Length of Opposite Side]/[Length of Hypotenuse]

We use this definition of sine to write an equation.

sin(55°) = [x/2]/[10]

Solve for x.

Cheers

 

[pka]

If the radius of the circle is 10 cm, and the inscribed angle is 55 degrees, find the length of the subtended chord.

For a complete discussion and list of formulae see this.

 

[Calculator]

I fixed some typos, in my first reply. Sorry for any confusion.


View attachment 4067

The rays of the inscribed angle are in blue, and the rays of the central angle are in red.

The subtended chord is shown as a thick, grey line segment. You're trying to find the length of the chord. You named this length x.

If we bisect the triangle with red sides (those side lengths are equal to the circle's radius, by the way), then both the 110° angle and the chord are cut in half and we get two identical right triangles. Use one of these right triangles.

Its hypotenuse is exactly the same as the radius of the circle.

The 110° angle got cut in half, so the corresponding angle inside the right triangle is 110°/2, or 55° (same as the blue, inscribed angle -- that always happens).

The length of the right-triangle's side that is opposite the (110°)/2 angle is x/2.

The right-triangle trigonometric ratio for sine is sin(angle)=[Length of Opposite Side]/[Length of Hypotenuse]

We use this definition of sine to write an equation.

sin(55°) = [x/2]/[10]

Solve for x.

Cheers

Thank you so much for your effort, I just needed to see the sketch but you even took your time to explain it to me thoroughly. Cheers :grin::grin: