# Two observers A and B are 100 ft apart and working at an object X above them. How far is object from observer A if angle B is 30 degrees and angle A is 45 degrees

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by rjay36

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by rjay36

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Angle from object X = 180° - (∠A + ∠B)

= 180° - (45° + 30°)

= 180° - 75°

=105°

∠X = 105°

x (side opposite X) = 100 ft.

∠A = 45°

a (side opposite A)= distance from B to X

∠B = 30°

b (side opposite B) = distance from A to X

Solve using Law of Sines:

1). Solve for a, or the distance from B to Object X.

x a

-------- = -------

sin X sin A

100 ft a

-------------- = ---------

sin 105° sin 45°

a (sin 105°) = 100 ft (sin 45°)

100 ft (sin 45°)

a = ----------------------

sin 105°

100 ft (0.707)

a = --------------------

0.966

70.7 ft

a = -------------

0.966

**a = 73.19 ft or 73 ft ⇒ distance from B to Object X. **

2.) Solve for b, or the distance from A to Object X:

100 ft (sin 30°)

b = -------------------

sin 105°

100 ft (0.5)

b = -----------------

0.966

50 ft

b = -----------

0.966

**b = 51.75 or 52 ft. ⇒ distance from A to Object X.**

**AN****SWER: The distance from observer A to Object X is 52 ft, while the distance from observer B to to Object X is 73 ft.**

= 180° - (45° + 30°)

= 180° - 75°

=105°

∠X = 105°

x (side opposite X) = 100 ft.

∠A = 45°

a (side opposite A)= distance from B to X

∠B = 30°

b (side opposite B) = distance from A to X

Solve using Law of Sines:

1). Solve for a, or the distance from B to Object X.

x a

-------- = -------

sin X sin A

100 ft a

-------------- = ---------

sin 105° sin 45°

a (sin 105°) = 100 ft (sin 45°)

100 ft (sin 45°)

a = ----------------------

sin 105°

100 ft (0.707)

a = --------------------

0.966

70.7 ft

a = -------------

0.966

2.) Solve for b, or the distance from A to Object X:

100 ft (sin 30°)

b = -------------------

sin 105°

100 ft (0.5)

b = -----------------

0.966

50 ft

b = -----------

0.966