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Vika [28.1K]
3 years ago
14

Giving brainlist to whoever answers

Mathematics
1 answer:
alexira [117]3 years ago
7 0
Correct answer is B
10/15=.666
This is RW divided by RT
Now you have to use the scale factor. Multiply by 18 to get WV
18 x 0.666 = 12

Good Luck Thnks for brainiest :)
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Two ships leave a harbor together, traveling on courses that have an angle of 135°40' between them. If they each travel 402 mile
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Answer:

Therefore they are 734.106 miles apart.

Step-by-step explanation:

Given that ,

Two ships have a harbor together. The angle between two ships  is  135°40'. Each of two ships travel 402 miles.

It forms a isosceles triangle whose two sides are 402 miles and one angle is 135°40'. Since it is isosceles triangle then other two angles of the triangle is equal.

Let ∠B= 135°40', and AB = 402 miles , BC =  402 miles

Then the distance between the ships = AC

We know

The sum of all angles = 180°

⇒∠A+∠B+∠C=180°

⇒∠A+135°40'+∠C=180°

⇒2∠A= 180°- 135°40'      [ since ∠A=∠C]

⇒2∠A=44°60'

⇒∠A= 22°30'

Again we know that,

\frac{AB}{sin\angle C}=\frac{BC}{sin \angle A}=\frac{AC}{sin \angle B}

Taking last two ratio,

\frac{BC}{sin \angle A}=\frac{AC}{sin \angle B}

Putting the value of BC , AC ,∠A,∠B

\frac{402}{sin 22^\circ30'}=\frac{AC}{sin 135^\circ40'}

\Rightarrow AC=\frac{402 \times sin135^\circ40'}{sin 22^\circ30'}

         ≈734.106 miles

Therefore they are 734.106 miles apart.

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