The bearings of ships A and B from a port P are 225° and 116° respectively. Ship A is 3.9km from ship B on a bearing of 258°.

Mathematics

1 answer:

Kobotan [32]3 years ago

5 0

Answer:

2.54 km

Step-by-step explanation:

<em>See the attached</em>

All given details are reflected in the diagram (not to scale)

PA is at 225°, PB is at 116°, BA is at 258°
AB = 3.9 km
AP = x

m∠P = 225° - 116° = 109°
m∠A = 258° - 225° = 33°
m∠B = 180° - (109° + 33°) = 38°

<u>Use the law of sines:</u>

x/ sin 38° = 3.9 / sin 109°
x = 3.9 sin 38° / sin 109°
x = 2.54 km (rounded)

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In the right ΔABC, AN is the altitude to the hypotenuse. Find BN, AN, and AC, if AB=2√5 in and NC= 1 in.

kipiarov [429]

Answer:

$BN=4\ \text{in}$

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Step-by-step explanation:

Let $BN=x$

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We have the relation

$\dfrac{BC}{AB}=\dfrac{AB}{BN}\\\Rightarrow \dfrac{c}{a}=\dfrac{a}{x}\\\Rightarrow c=\dfrac{a^2}{x}\\\Rightarrow x+1=\dfrac{a^2}{x}\\\Rightarrow x^2+x=20\\\Rightarrow x^2+x-20=0\\\Rightarrow x=\frac{-1\pm \sqrt{1^2-4\times 1\times \left(-20\right)}}{2\times 1}\\\Rightarrow x=4,-5$