Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 7 - Applications of Trigonometry and Vectors - Section 7.5 Applications of Vectors - 7.5 Exercises - Page 342: 17

Answer

$v\approx\left\langle -123.255,154.955 \right\rangle$

Work Step by Step

The horizontal component of a vector $v=\left\langle a,b \right\rangle$ having magnitude $|v|$ and 1direction angle $\theta$ is given by $a=|v|\cos\theta$. Similarly, the vertical component is given by $b=|v|\sin\theta$. Now we have $|v|=198$ and $\theta=128.5^{\circ}$ therefore, the horizontal component is \begin{align*} a=&|v|\cos\theta\\ a=&198\cos128.5^{\circ}~~~ \text{use} ~~\theta=128.5^{\circ} ,|v|=198\\ a=&198\times (-0.6225)\\ a=&-123.255~~~(\text{approximated to three decimal places}), \end{align*} and the vertical component is \begin{align*} b=&|v|\sin\theta\\ b=&198\sin128.5^{\circ}~~~ \text{use} ~~\theta=128.5^{\circ} ,|v|=198\\ b=&198\times 0.7826\\ b =&154.955~~~~~~ (\text{approximated to three decimal places}). \end{align*} Hence, the vector $v=\left\langle a,b \right\rangle\approx\left\langle -123.255,154.955 \right\rangle$ .
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