document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1169043>Question 1169043</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Using trigonometry determine the resultant vector\r<BR>\n" );
document.write( "\n" );
document.write( "a) 8m [N] and 12 m [W]\r<BR>\n" );
document.write( "\n" );
document.write( "b) 15N [30degreesSofE] and 8 N [E] </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #851457 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1959)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=851457\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let's determine the resultant vector for each case using trigonometry.\r<BR>\n" );
document.write( "\n" );
document.write( "**a) 8 m [N] and 12 m [W]**\r<BR>\n" );
document.write( "\n" );
document.write( "1.  **Represent the vectors in component form:**<BR>\n" );
document.write( "    * Vector 1 (A): 8 m [N] lies along the positive y-axis.<BR>\n" );
document.write( "        $\mathbf{A} = (0, 8)$<BR>\n" );
document.write( "    * Vector 2 (B): 12 m [W] lies along the negative x-axis.<BR>\n" );
document.write( "        $\mathbf{B} = (-12, 0)$\r<BR>\n" );
document.write( "\n" );
document.write( "2.  **Add the component vectors to find the resultant vector (R):**<BR>\n" );
document.write( "    $\mathbf{R} = \mathbf{A} + \mathbf{B} = (0 + (-12), 8 + 0) = (-12, 8)$\r<BR>\n" );
document.write( "\n" );
document.write( "3.  **Calculate the magnitude of the resultant vector:**<BR>\n" );
document.write( "    $|\mathbf{R}| = \sqrt{R_x^2 + R_y^2} = \sqrt{(-12)^2 + (8)^2} = \sqrt{144 + 64} = \sqrt{208}$<BR>\n" );
document.write( "    $|\mathbf{R}| = \sqrt{16 \times 13} = 4\sqrt{13} \approx 14.42$ m\r<BR>\n" );
document.write( "\n" );
document.write( "4.  **Calculate the direction of the resultant vector:**<BR>\n" );
document.write( "    The angle $\theta$ that the resultant vector makes with the positive x-axis can be found using the arctangent function:<BR>\n" );
document.write( "    $\tan(\theta) = \frac{R_y}{R_x} = \frac{8}{-12} = -\frac{2}{3}$<BR>\n" );
document.write( "    $\theta = \arctan(-\frac{2}{3}) \approx -33.69^\circ$\r<BR>\n" );
document.write( "\n" );
document.write( "    Since $R_x$ is negative and $R_y$ is positive, the resultant vector lies in the second quadrant. To find the angle relative to the positive x-axis, we add $180^\circ$:<BR>\n" );
document.write( "    $\theta_{actual} = -33.69^\circ + 180^\circ = 146.31^\circ$\r<BR>\n" );
document.write( "\n" );
document.write( "    To express the direction in a standard compass bearing format, we can find the angle relative to the North or West direction. The angle with the negative x-axis (West) is $33.69^\circ$. Therefore, the direction is approximately $33.69^\circ$ North of West [N $33.69^\circ$ W].\r<BR>\n" );
document.write( "\n" );
document.write( "    Alternatively, the angle with the positive y-axis (North) is $90^\circ - 33.69^\circ = 56.31^\circ$. Therefore, the direction is approximately $56.31^\circ$ West of North [W $56.31^\circ$ N].\r<BR>\n" );
document.write( "\n" );
document.write( "    Both [N $33.69^\circ$ W] and [W $56.31^\circ$ N] are correct ways to express the direction.\r<BR>\n" );
document.write( "\n" );
document.write( "**Resultant vector for a): Approximately 14.42 m [N 33.69° W]**\r<BR>\n" );
document.write( "\n" );
document.write( "**b) 15 N [30° S of E] and 8 N [E]**\r<BR>\n" );
document.write( "\n" );
document.write( "1.  **Represent the vectors in component form:**<BR>\n" );
document.write( "    * Vector 1 (A): 15 N [30° S of E]<BR>\n" );
document.write( "        $A_x = |\mathbf{A}| \cos(30^\circ) = 15 \times \frac{\sqrt{3}}{2} \approx 15 \times 0.866 = 12.99$ N (East)<BR>\n" );
document.write( "        $A_y = -|\mathbf{A}| \sin(30^\circ) = -15 \times \frac{1}{2} = -7.5$ N (South)<BR>\n" );
document.write( "        $\mathbf{A} = (12.99, -7.5)$<BR>\n" );
document.write( "    * Vector 2 (B): 8 N [E] lies along the positive x-axis.<BR>\n" );
document.write( "        $\mathbf{B} = (8, 0)$\r<BR>\n" );
document.write( "\n" );
document.write( "2.  **Add the component vectors to find the resultant vector (R):**<BR>\n" );
document.write( "    $\mathbf{R} = \mathbf{A} + \mathbf{B} = (12.99 + 8, -7.5 + 0) = (20.99, -7.5)$\r<BR>\n" );
document.write( "\n" );
document.write( "3.  **Calculate the magnitude of the resultant vector:**<BR>\n" );
document.write( "    $|\mathbf{R}| = \sqrt{R_x^2 + R_y^2} = \sqrt{(20.99)^2 + (-7.5)^2} = \sqrt{440.5801 + 56.25} = \sqrt{496.8301} \approx 22.29$ N\r<BR>\n" );
document.write( "\n" );
document.write( "4.  **Calculate the direction of the resultant vector:**<BR>\n" );
document.write( "    The angle $\theta$ that the resultant vector makes with the positive x-axis (East) can be found using the arctangent function:<BR>\n" );
document.write( "    $\tan(\theta) = \frac{R_y}{R_x} = \frac{-7.5}{20.99} \approx -0.3573$<BR>\n" );
document.write( "    $\theta = \arctan(-0.3573) \approx -19.67^\circ$\r<BR>\n" );
document.write( "\n" );
document.write( "    The negative angle indicates that the resultant vector is below the positive x-axis. Therefore, the direction is approximately $19.67^\circ$ South of East [S $19.67^\circ$ E].\r<BR>\n" );
document.write( "\n" );
document.write( "**Resultant vector for b): Approximately 22.29 N [S 19.67° E]** </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );