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Trigonometry Triangle and circle length and areas

karush

Well-known member
Jan 31, 2012
2,723
the following diagram shows a circle with center \(\displaystyle O\) and a radius
\(\displaystyle 4cm\)
View attachment 1006



The points \(\displaystyle A, B,\) and \(\displaystyle C\) Lie on the circle.
The point \(\displaystyle D\) is outside the circle, on \(\displaystyle (OC)\)
Angle \(\displaystyle ADC=0.3\) radians and angle \(\displaystyle AOC=0.8\) radians

(a) find \(\displaystyle AD\)

I used law of sines

\(\displaystyle \frac{4}{\sin{0.3}}=\frac{x}{\sin{0.8}}\)
\(\displaystyle x \approx 9.71cm\)

there are more questions to this but want to make sure this is correct:cool:
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: triangle and circle length and areas

Yes, you have correctly applied the Law of Sines. (Sun)
 

karush

Well-known member
Jan 31, 2012
2,723
Re: triangle and circle length and areas

(b) find OD

since \(\displaystyle \angle{DAO}\) is not given the its radian measure is
\(\displaystyle 2-0.3-0.8=0.9\)

so using law of sines again

\(\displaystyle \frac{x}{\sin(0.9)}=\frac{4}{\sin(0.3)}\)

so \(\displaystyle x \approx 10.6\)

 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: triangle and circle length and areas

The sum of the interior angles of a triangle in radians is $\pi$, not $2$.

Also, when you go to use this, be aware of the identity $\sin(\pi-\theta)=\sin(\theta)$.
 

karush

Well-known member
Jan 31, 2012
2,723
Re: triangle and circle length and areas

\(\displaystyle \pi-1.1 \approx 2.04159\)

\(\displaystyle \frac{x}{sin(2.4159)}=\frac{4}{sin(0.3)}\)

\(\displaystyle x \approx 13.53\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: triangle and circle length and areas

You appear to have dropped a zero to the right of the decimal point in the argument for the sine function on the left. I would write:

\(\displaystyle \frac{\overline{OD}}{\sin(A)}=\frac{4\text{ cm}}{\sin(0.3)}\)

Now, given:

\(\displaystyle 0.3+0.8+A=\pi\,\therefore\,A=\pi-1.1\)

and using the identity \(\displaystyle \sin(\pi-\theta)=\sin(\theta)\)

we have:

\(\displaystyle \frac{\overline{OD}}{\sin(1.1)}=\frac{4\text{ cm}}{\sin(0.3)}\)

\(\displaystyle \overline{OD}=\frac{(4\text{ cm})\sin(1.1)}{\sin(0.3)}\approx12.062895734\text{ cm}\)
 

karush

Well-known member
Jan 31, 2012
2,723
View attachment 1009

(c) find the area of sector \(\displaystyle OABC\)

\(\displaystyle \Bigg(\frac{0.8}{2\pi}\Bigg)\Bigg(\pi 4^2 \Bigg)\approx 6.4 cm^2\)

last question

(d) Find the area of region \(\displaystyle ABCD\)

\(\displaystyle \frac{1}{2} (12.0624)(4\sin{0.8})-6.4 \approx 10.91 cm^2\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
c) The area $A$ of a circular sector having radius $r$ and subtending an angle $\theta$ is given by:

\(\displaystyle A=\frac{1}{2}r^2\theta\)

In this case what are $r$ and $\theta$?
 

karush

Well-known member
Jan 31, 2012
2,723
c) The area $A$ of a circular sector having radius $r$ and subtending an angle $\theta$ is given by:

\(\displaystyle A=\frac{1}{2}r^2\theta\)

In this case what are $r$ and $\theta$?
r=4 and \theta = 0.8

so

\(\displaystyle A=\frac{1}{2} 4^2 \ (0.8)=6.4 cm\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
The magnitude of the result is correct, but the unit of area is not, which may seem very minor now, but if you take physics, keeping track of the units becomes important, and it is a good habit to get into early on. I would write:

\(\displaystyle A=\frac{1}{2}(4\text{ cm})^2\cdot0.8=6.4\text{ cm}^2\)

You should expect an area to have as its unit of measure the square of a linear measure.
 

karush

Well-known member
Jan 31, 2012
2,723
looks like just need to be more careful:)

thanks again for help