5.6.4 Exercises

What is the degree measure of the angle $\phi$ between the $x$-axis and the connecting line from the origin in the Cartesian coordinate system to the point $P_{\phi }=(-0.643;-0.766)$ on the unit circle? Use a calculator, but do not trust it blindly!
Result: $\phi$$\hspace{0.5em}=\hspace{0.5em}$
${}^{\circ }$
Solution

From the coordinates of the point $P_{\phi }$ we have

${\displaystyle \cos \left(\alpha \right)=-0.643\mathrm{ }\text{and}\mathrm{ }\sin \left(\alpha \right)=-0.766\hspace{0.5em}.}$
If you enter

• invers(cos(-0.643)) or ${\cos }^{-1}$(-0.643) in the calculator, you obtain approximately $130{}^{\circ }$
  
• invers(sin(-0.766)) or ${\sin }^{-1}$(-0.766) in the calculator, you obtain approximately $-50{}^{\circ }$.
  

Moreover, you know that the point lies in the third quadrant. Thus, the angle must be in the range from $180{}^{\circ }$ to $270{}^{\circ }$.

The figure to the left shows that the negative cosine value corresponds to the angle $-130{}^{\circ }$ and to the angle $\phi =-130{}^{\circ }=-130{}^{\circ }+360{}^{\circ }=230{}^{\circ }$. Likewise, the negative sine value can correspond to the angle $-50{}^{\circ }$ and to the angle $\phi =-(-50{}^{\circ })+180{}^{\circ }=230{}^{\circ }$. Since this last value lies in the range stated above, the required value of the angle is $\phi =230{}^{\circ }$ indicated in the figure by a pink line.

1. Let a right triangle with the right angle at the vertex $C$ and the sides $b=2.53\hspace{0.17em}\mathrm{cm}$ and $c=3.88\hspace{0.17em}\mathrm{cm}$ be given. Calculate the values of $\sin \left(\alpha \right)$, $\sin \left(\beta \right)$, and $a$.
   Results:
   • $\sin \left(\alpha \right)$$\hspace{0.5em}=\hspace{0.5em}$
     
   • $\sin \left(\beta \right)$$\hspace{0.5em}=\hspace{0.5em}$
     
   • $a$$\hspace{0.5em}=\hspace{0.5em}$
     $\hspace{0.17em}\mathrm{cm}$
     
   
   Solution
   We have
   
   ${\displaystyle a=\sqrt{c^2-b^2}=\sqrt{{\left(3.88\hspace{0.17em}\mathrm{cm}\right)}^2-{\left(2.53\hspace{0.17em}\mathrm{cm}\right)}^2}=\sqrt{15.0544\hspace{0.17em}\mathrm{cm}{}^2-6.4009\hspace{0.17em}\mathrm{cm}{}^2}=\sqrt{8.6535}\hspace{0.17em}\mathrm{cm}\hspace{0.5em},}$
   and
   
   ${\displaystyle \sin \left(\alpha \right)=\frac{a}{c}=\frac{\sqrt{8.6535}\hspace{0.17em}\mathrm{cm}}{3.88\hspace{0.17em}\mathrm{cm}}=\frac{\sqrt{86535}}{388}\mathrm{ }\text{and}\mathrm{ }\sin \left(\beta \right)=\frac{b}{c}=\frac{2.53\hspace{0.17em}\mathrm{cm}}{3.88\hspace{0.17em}\mathrm{cm}}=\frac{253}{388}\hspace{0.5em}.}$
   Numerically, we obtain $a\approx 2.9417\hspace{0.17em}\mathrm{cm}$, $\sin \left(\alpha \right)\approx 0.7587$, and $\sin \left(\beta \right)\approx 0.65201$.
   
   
2. Calculate the area $F$ of a triangle with the sides $a=4\hspace{0.17em}\mathrm{m}$, $c=60\hspace{0.17em}\mathrm{cm}$, and the angle $\beta =\measuredangle (a,c)=\frac{11\pi }{36}$.
   Result: $F$$\hspace{0.5em}=\hspace{0.5em}$
   $\hspace{0.17em}\mathrm{m}{}^2$
   Please, enter your result rounded to three decimal digits or as an expression. Enter the sine of an angle as sin(x) and the number $\pi$ as pi.
   Solution
   
   
   ${\displaystyle \frac{(a·\sin \left(\beta \right))·c}{2}=\sin \left(\frac{11\pi }{36}\right)·1.2\hspace{0.17em}\mathrm{m}{}^2\approx 0.98298\hspace{0.17em}\mathrm{m}{}^2\hspace{0.5em}.}$