Question

Two points A and B are on opposite sides of a building. A surveyor selects a third point C to place a transit. Point C is 52 feet from point A and 64 feet from point B. The angle ACB is 52°. How far apart are points A and B? Round to the 10th place.

Answer 1

Answer:

Step-by-step explanation:1.

Use law of cosine.

AB^2 = AC^2 + BC^2 - 2 * AC * BC * cos(angle ACB)

AB^2 = 53^2 + 71^2 - 2(53)(71)cos(53)

Using calculator,

AB = about 57.6259 feet

2.

Law of cosine again.

d^2 = 400^2 + 90^2 - 2(400)(90)cos(45)

using calculator,

d = about 342.3278 feet

Answer 2

The distance between points A and B (or AB) is equal to 51.98 feet.

Given information:

Point C is 52 feet from point A (AC=52), and 64 feet from point B (BC=64).

The angle ACB is 52°.

It is required to find the distance between A and B (AB).

See the attached image.

Use the cosine law in the triangle ABC to find the value of AB as,

$AB^2=AC^2+BC^2-2AB\times BC \times cos\angle ACB\\AB^2=52^2+64^2-252\times 64\times cos 52\\AB^2=2702.15\\AB=51.98$

The value of side AB is 52.98 feet.

Therefore, the distance between points A and B (or AB) is equal to 51.98 feet.

For more details, refer to the link:

https://brainly.com/question/21634338