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wel
3 years ago
9

What is the approximate value of b, rounded to the nearest tenth? Use the law of sines to find the answer.

Mathematics
2 answers:
vivado [14]3 years ago
3 0

You can use the Law of Sines for this problem, but what makes it tough at first glance is that they put a step in before you can use it directly. That first step is to get m∠C. When solving a triangle, always get the third angle if you know the other two so that you can use the Law of Sines (or Law of Cosines) more directly.


The sum of the angles of any triangles is 180°. We know the measures of the two other angles. So we subtract them from 180° to find that third angle.


180° - 76° - 66° = 38°


Now we can use the proportion from the picture. Angle B measures 76°, side b is our unknown, Angle C measures 38°, and side c measures 3 units.


sin 76° sin 38°

----------- = -----------

b 3


Cross multiply to find b (the side). We carry out five places in these calculations to get more accuracy.


3 * sin 76° = b * sin 38°


b = 3 * sin 76° 3 * 0.56611 1.69832

--------------- = --------------- = ---------------- = 5.73044

sin 38° 0.29636 0.29636


To the nearest tenth, b - 5.7 units.

Alla [95]3 years ago
3 0

Answer:

just took the quiz 4.7 is the answer

Step-by-step explanation:

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Use the alternative form of the dot product to find u · v.
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Answer:

Step-by-step explanation:

u = 45

v = 30

Angle between them, θ = 5π / 6 = 150°

(a) The formula for the dot product is given by

\overrightarrow{u}.\overrightarrow{v}= u v Cos\theta

\overrightarrow{u}.\overrightarrow{v}= 45 \times 30 \times Cos150

\overrightarrow{u}.\overrightarrow{v}= -1169.13

(b) \overrightarrow{u}=Cos\frac{\pi }{3}\widehat{i}+Sin\frac{\pi }{3}\widehat{j}

\overrightarrow{u}=0.5\widehat{i}+0.866\widehat{j}

\overrightarrow{v}=Cos\frac{2\pi }{3}\widehat{i}+Sin\frac{2\pi }{3}\widehat{j}

\overrightarrow{v}=-0.707\widehat{i}+0.707\widehat{j}

Let the angle between them is θ

The formula in terms of the dot product is given by

Cos\theta =\frac{\overrightarrow{u}.\overrightarrow{v}}{u v}

u=\sqrt{0.5^{2}+0.866^{2}}=1

v=\sqrt{-0.707^{2}+0.707^{2}}=1

Cos\theta =\frac{0.5 \times (-0.707) + 0.866 \times 0.707)}{1\times 1}

Cos\theta=0.2587

θ = 75°

(c) \overrightarrow{u}=1\widehat{i}+4\widehat{j}+8\widehat{k}

u=\sqrt{1^{2}+4^{2}+8^{2}}=9

Cos\alpha =\frac{1}{9}

Cos\beta =\frac{4}{9}

Cos\gamma  =\frac{8}{9}

7 0
3 years ago
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