솔루션피아

$$ \begin{cases} \vec d_1+\vec d_2&=6\vec d_3\\ \vec d_1-\vec d_2&=4\vec d_3\\ \vec d_3&=3\i+5\j\\ \end{cases} $$ $$ \begin{cases} \vec d_1&=5\vec d_3\\ \vec d_2&=\vec d_3\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} \vec d_1&=5\vec d_3\\ &=5\(3\i+5\j\)\\ &=15\i+25\j\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \vec d_2&=\vec d_3\\ &=3\i+5\j\\ \end{aligned} $$

$$ \begin{cases} \vec a&=7.00\ut{m}\i\\ b&=5.00\ut{m},\theta_b=90\degree+35\degree\\ \end{cases} $$ $$ \begin{aligned} \vec b &= 5\cos125\degree\i+5\sin125\degree\j\\ \end{aligned} $$ $$\ab{a,b}$$ $$ \begin{aligned} \vec c &= \vec a+\vec b\\ &=\(7-5 \sin35\degree\)\i+5 \cos35\degree\j\\ \end{aligned} $$ $$\ab{a}$$ $$ \begin{aligned} c&=\sqrt{\(7-5 \sin35\degree\)^2+(5\cos35\degree)^2}\\ &=\sqrt{..

$$ \begin{cases} \vec a &= 3.0\i+4.0\j-3.0\k\\ \vec b &= -2.0\i+2.0\j+3.0\k\\ \vec c &= 4.0\i+3.0\j+4.0\k\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} \vec r &= \vec a-\vec b+\vec c\\ &=9.0\i+5.0\j-2.0\k\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \theta_{r~z+}&=\cos^{-1}\frac{r_z}{r}\\ &=\cos^{-1}\frac{-2}{\sqrt{9^2+5^2+(-2)^2}}\\ &=\cos ^{-1}\left(-\sqrt{\frac{2}{55}}\right)\\ &\approx ..

$$ \begin{cases} d_1&=6.00\ut{m},\theta_{1}=-135\degree\\ \vec d_2&=4.00\ut{m}\i\\ d_3&=8.00\ut{m},\theta_{3}=60.0\degree\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} d_{1x}&=6\cos\(-135\degree\)\\ &=-3\sqrt2\ut{m}\\ &\approx -4.242640687119286\ut{m}\\ &\approx -4.24\ut{m} \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} d_{1y}&=6\sin\(-135\degree\)\\ &=-3\sqrt2\ut{m}\\ &\approx -4.2426406871192..

$$ \begin{cases} \vec a - \vec b &= 3\vec c\\ \vec a + \vec b &= 5\vec c\\ \vec c &=6\i+4\j\\ \end{cases} $$ $$ \begin{cases} \vec a &= 4\vec c\\ \vec b &= \vec c\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} \vec a &= 4\vec c\\ &= 4\(6\i+4\j\)\\ &= 24\i+16\j \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \vec b &= \vec c\\ &= 6\i+4\j\\ \end{aligned} $$

$$ \begin{cases} a&=4.30\\ b&=6.20\\ \phi&=50.0\degree\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} A&=\frac{1}{2}ab\sin\phi\\ &=\frac{1}{2}\cdot4.3\cdot6.2\sin50\degree\\ &\approx 10.21137242677598\\ &\approx 10.2\\ \end{aligned} $$ $$\ab{b}$$ $$\vec a\times \vec b =ab\sin\phi,$$ $$ \begin{aligned} A&=\frac{1}{2}ab\sin\phi\\ &=\frac{1}{2}\(\vec a\times \vec b\)\\ \end{aligned} $$

$$ \begin{cases} d_1&=0.30\ut{m},\theta_{1~y-}=-45\degree\\ \vec d_2&=0.250\ut{m}\i\\ d_3&=0.60\ut{m},\theta_{3~x+}=60.0\degree\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} d_{1x}&=0.3\cos\(-90\degree-45\degree\)\\ &=-\frac{3}{10 \sqrt{2}}\ut{m}\\ &\approx -0.2121320343559642\ut{m}\\ &\approx -0.21\ut{m} \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} d_{1y}&=0.3\sin\(-90\degree-45\degree\)\\ &..

$$ \begin{cases} P &= 13.0\ut{m}, \theta_P=25.0\degree\\ Q &= 10.0\ut{m}, \theta_Q=(90.0+10.0)\degree\\ R &= 9.00\ut{m}, \theta_R=(90.0-20.0)\degree\\ S &= 8.00\ut{m}, \theta_R=(-90.0+40.0)\degree\\ \end{cases} $$ $$ \begin{cases} \vec P &= 13\cos25\degree\i+13\sin25\degree\j\\ \vec Q &= 10\cos100\degree\i+10\sin100\degree\j\\ \vec R &= 9\cos70\degree\i+9\sin70\degree\j\\ \vec S &= 8\cos(-50\deg..

$$ \begin{cases} \vec d_1&=3.0\i+4.0\j-7.0\k\\ \vec d_2&=-1.0\i-4.0\j+3.0\k\\ \vec d_3&=-5.0\i+3.0\j+2.0\k\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} \vec r&=\vec d_1-\vec d_2+\vec d_3\\ &=-i+11\j-8\k\\ &=-1.0\i+11\j-8.0\k\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \theta_{a~z+}&=\cos^{-1}\frac{r_z}{r}\\ &=\cos^{-1}\frac{-8}{\sqrt{(-1)^2+11^2+(-8)^2}}\\ &=\cos ^{-1}\left(-4 \sqrt{\frac..

$$ \begin{cases} A&=18.0\ut{m}\\ \theta_A&=60\degree\\ \vec B&=12.0\ut{m}\i+8.00\ut{m}\j\\ \phi &= 30.0\degree\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} A_2&=A_1=18\\ \theta_{A2}&=\theta_{A1}-\phi\\ &=60\degree-30\degree\\ &=30\degree \end{aligned} $$ $$ \begin{aligned} \vec A_2&=18\cos30\degree\i+18\sin30\degree\j\\ &=\(9\sqrt3\i+9\j\)\ut{m}\\ &\approx \(15.58845726811989\i+9\j\)\ut{m}\\ &..