Looking for answers? Westonci.ca is your go-to Q&A platform, offering quick, trustworthy responses from a community of experts. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
Since all the sides of the figure have the same length, then the figure is a rhombus. Then, its diagonals intersect at an angle of 90°.
Let O be the intersection of the diagonals of the rhombus. Notice that the triangle EOA is a right triangle. Since the side EA is the hypotenuse of the triangle, then, recalling the trigonometric functions:
[tex]\begin{gathered} \cos (30)=\frac{EO}{EA} \\ \sin (30)=\frac{OA}{EA} \end{gathered}[/tex]Use this information to solve for the segments EO and OA:
[tex]\begin{gathered} EO=EA\cdot\cos (30) \\ =10\cdot\frac{\sqrt[]{3}}{2} \\ =5\cdot\sqrt[]{3} \end{gathered}[/tex][tex]\begin{gathered} OA=EA\cdot\sin (30) \\ =10\cdot\frac{1}{2} \\ =5 \end{gathered}[/tex]Since the diagonal EM is twice the segment EO and the diagonal BA is twice the segment OA, then the lengths of the diagonals are:
[tex]\begin{gathered} BA=10 \\ EM=10\cdot\sqrt[]{3} \end{gathered}[/tex]Therefore, the answer is:
[tex]10\text{ yards and }10\cdot\sqrt[]{3}\text{ yards}[/tex]
Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.
