Question

1.
Berkley is flying a kite. The string is all the way out, which means it is 425 meters away. Berkley is looking up at the kite at an angle of 42°. Berkley's dog is watching the kite too and the angle from Berkley to the dog to the kite is 87°. How would you find the distance between the kite and the dog? Is it possible? Explain your answer using the law of sines.

Answer 1

Using the law of sines, it is found that the distance between the kite and the dog is of 284.77 meters.

What is the law of sines?

Suppose we have a triangle in which:

• The length of the side opposite to angle A is a.

• The length of the side opposite to angle B is b.

• The length of the side opposite to angle C is c.

The lengths and the sine of the angles are related as follows:

$\frac{\sin{A}}{a} = \frac{\sin{B}}{b} = \frac{\sin{C}}{c}$

For the situation described, we have that:

• The height is opposite to the angle of 42º.

• The 425 meters are opposite to the angle of 87º.

Hence:

$\frac{\sin{42^\circ}}{h} = \frac{\sin{87^\circ}}{425}$

Applying cross multiplication:

$h = 425\frac{\sin{42^\circ}}{\sin{87^\circ}}$

h = 284.77 meters.

More can be learned about the law of sines at https://brainly.com/question/25535771

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