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3 years ago

Cos155° = _____ -cos25° cos 55° cos(-25)°

Mathematics

1 answer:

Sphinxa [80]3 years ago

3 0

Answer:

- cos 25°

Step-by-step explanation:

Cosine function is one of the trigonometric functions. Cosine function is regarded as an even function, which means that f(-x) = f(x). Also, cosine function is positive in the first quadrant and the last quadrant and negative in the second quadrant and the third quadrant. 155° lies in the second quadrant since 155° is smaller than 180°. Therefore, the basic angle or the reference angle of 155° is 180° - 155° = 25°. We know that cos 155° will be negative because it lies in the second quadrant and cos 25° will be positive because it lies in the first quadrant. Since cos 55° is positive, and cos (-25°) = cos 25° by the even function property, therefore option 2 and option 3 are incorrect since cos 155° is negative. Therefore, option 1 is the correct answer i.e. cos 155° = - cos 25°!!!

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What is the GFC of 40 and 60

Svetradugi [14.3K]

Answer:

GCF: 20

Step-by-step explanation:

The thing you want to do is factor both numbers down to their lowest primes.

40:2 * 2 * 2 * 5

60:2 * 2 * 5 * 3

Now underline what is in both sets of primes. I'll bold them. The number that is bolded comes from 2 * 2 * 5 which is 20

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3 years ago

NGC is a hot air balloon in the sky from her spot on the ground. The angle of elevation from Angie to the balloon is 40°. If she

Goshia [24]

Answer:

The distance from hot air balloon to the ground $AC=50.1457 ft$ .

Step-by-step explanation:

Labelled diagram of given scenario is shown below.

Given that,

An angle of elevation of Hot air balloon by Angie is $40$ °.

When she stepped back $200 ft$ then angle of elevation was $10$ °.

Height of Angie is $5.5 ft$ .

To find: How far off the ground is a hot air balloon.

So, from figure

Height of Angie $(BC) = 5.5ft$

In triangle Δ $ABD$ ,

⇒ $tan {40\si {\degree}} = \frac{AB}{BD}$

⇒ $AB = tan (40) \times BD$ ....................(1)

Now, In triangle Δ $ABE$

⇒ $tan (10)= \frac{AB}{200+BD}$

⇒ $AB = tan(10)\times (200+BD)$

Here, substituting the value $AB$ from Equation (1) we get,

$tan(10)\times (200+BD)= tan(40)\times BD$

$tan(10)\times 200+tan(10)\times BD= tan(40)\times BD$

⇒ $BD\times (tan(40)-tan(10))=tan(10)\times 20$

⇒ $BD\times 0.6628 = 35.2654$

⇒ $BD = \frac{35.2654}{0.6628} =53.2067 ft$

Now, finding the value of $AB$ from equation (1)

$AB= tan(40)\times 53.2067=0.8391\times 53.2067$

$AB=44.6457 ft$

Therefore Length of $AC= AB+BC$ = $(44.6457 + 5.5) ft$

$AC=50.1457 ft$ .

Hence,

The distance from hot air balloon to the ground $AC=50.1457 ft$ .

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