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2 years ago

A pile of sand has a weight of 90kg the sand is put into a small bag and a medium bag and a large bag in the ratio 2:3:7

Mathematics

1 answer:

Finger [1]2 years ago

3 0

90kg of sand had been puted in 3 bags but in 2+3+7=12 parts.
90kg/12=7.5kg
Small bag has 2 parts which is 7.5kg*2=15kg
Medium bag has 3 parts which is 7.5kg*3=22.5kg
Large bag has 7 parts which is 7.5kg*7=52.5kg
2:3:7=15kg:22.5kg:52.5kg
15kg+22.5kg+52.5kg=90kg

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Answer:

The value is $k =109.6 \ miles$

Step-by-step explanation:

The diagram illustrating the question is shown on the first uploaded image

From the question we are told that

The distance from city A to B is AB = 467.3 miles

The bearing from B to C is $\theta_{BC} = N 28.7E$

The bearing from B to A is $\theta_{BA} = N 60.7E$

The bearing from A to B is $\theta_{AB} = S60.7W$

The bearing from A to C is $\theta_{AC} = S79.1W$

Generally from the diagram

$\theta_A = 180 - 60.7 -79.1$

=> $\theta_A = 40.2 ^o$

Also

$\theta_B = 32^o$

and

$\theta_C = 180 - (\theta_A +\theta_B )$

=> $\theta_C = 180 - (40.2 + 32 )$

=> $\theta_C = 107.8 ^o$

Generally according to Sine Rule

$\frac{BC}{sin (\theta_A)} = \frac{CA}{sin (\theta_B)} =\frac{AB}{sin (\theta_C)}$

=> $\frac{BC}{sin (40.2)} = \frac{CA}{sin (32)} =\frac{467.3 }{sin (107.8)}$

$\frac{BC}{sin (40.2)} = \frac{467.3 }{sin (107.8)}$

=> $BC = 316.8 \ miles$

Also

$\frac{CA}{sin (32)} = \frac{467.3 }{sin (107.8)}$

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Generally the additional flyer miles that Adam will receive if he takes the connecting flight rather than the direct flight is mathematically represented as

$k = [CA +BC] - AB$

=> $k = [260 .1 +316.8]- 467.3$

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