Question #4dd1c

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Question #4dd1c

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Answer 1 · 2017-03-26T20:48:39

c

Explanation:

Draw 2 triangles from the vectors
vector a is 12m @ 40 degrees
vector b is 9m @ 20 degrees

Any line that's not straight can be a hypotnuse.
Based on the graph, you need cosine to find x since cos=hyp/adj and adj is moving in the x axis.

$cos θ = \frac{a d j}{h y p}$ $costheta=(adj)/(hyp)$ ---> $h y p cos θ = o p p$ $hypcostheta=opp$

Vector a: $a = 12 cos 40 = 9.19$ $a=12cos40=9.19$
Vector b: $b = 9 cos 20 = 8.46$ $b=9cos20=8.46$

a goes in the positive direction and b goes in the negative direction, so $9.19 - 8.46 = .735$ $9.19-8.46=.735$ which is basically c

Answer 2 · 2017-03-27T04:11:58

As given in the question

$\to a$ $veca$ subtends $40^{\circ}$ $40^@$ angle with +ve direction of X-axis and $∣ ∣ \to a ∣ ∣ = 12 m$ $absveca=12m$

$\to b$ $vecb$ subtends ${(180 - 20)}^{\circ} = 160^{\circ}$ $(180-20)^@=160^@$ angle with +ve direction of X-axis and $∣ ∣ ∣ \to b ∣ ∣ ∣ = 9 m$ $absvecb=9m$

So the component of $\to a + \to b$ $veca+vecb$ along X-axis will have magnitude

$= ∣ ∣ \to a ∣ ∣ {cos 40}^{\circ} + ∣ ∣ ∣ \to b ∣ ∣ ∣ {cos 160}^{\circ}$ $=absvecacos40^@+absvecbcos160^@$

$= (12 {cos 40}^{\circ} + 9 {cos 160}^{\circ}) m \approx 0.73 m$ $=(12cos40^@+9cos160^@)m~~0.73m$

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Question #4dd1c

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