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# can someone please answer both or 1 of these questions, thx!?

At night, a security camera pans over a parking lot. The camera is on

a post at point A, which is 35m from point C and 51m from point B. The

distance from B to C is 65m. Calculate ∠A, the angle through which the camera

pans, correct to the nearest degree.

Micah is standing on the ground between two buildings on the opposite sides of

a park. The base of the first building is 152 m from Micah, has an angle of

elevation of 38° to the top, while the base of the second building is 175 m from

Micah, and hast an angle of elevation of 53° to the top. How far apart are the

tops of the two buildings? Round your answer to the nearest meter.

### 2 Answers

- ♥Astrid♥Lv 63 weeks agoFavorite Answer
(1)

It doesn't explicitly state, but assuming the camera pans from B to C.

The Law of Cosines:

a² = b² + c² − 2bc cos(A)

cos(A) = (-a² + b² + c²) / 2bc

cos(A) = (-65² + 35² + 51²) / 2(35)(51)

A = 96°

(2)

Height of Bldg 1 = 152•tan38°

= 118.7m

Height of Bldg 2 = 175•tan53°

= 232.2m

Difference in heights = 232.2 - 118.7 = 113.5m

The diagonal from rooftop 1 to rooftop 2:

d = √((152 + 175)² + 113.5²)

d = 346m

- llafferLv 73 weeks ago
Sketching it out we get a triangle with three known lengths and no known angles. You want the measurement of angle "A", so we use its opposite length of 65 m in the law of cosines equation:

a² = b² + c² - 2bc cos(A)

65² = 35² + 51² - 2(35)(51) cos(A)

4225 = 1225 + 2601 - 3570 cos(A)

4225 = 3826 - 3570 cos(A)

399 = -3570 cos(A)

-399/3570 = cos(A)

A ≈ 96° (rounded to the nearest whole degree)