Observatories A,B
The target C is observed from two artillery observatories, A and B, 296 m apart. At the same time, angle BAC = 52°42" and angle ABC = 44°56". Calculate the distance of the target C from observatory A.
Correct answer:

Showing 1 comment:
Dr. Math
To determine the distance of target C from observatory A , we can use the Law of Sines in triangle ABC . Here's the step-by-step solution:
- The distance between observatories A and B is AB = 296 m .
- Angle BAC = 52°42' .
- Angle ABC = 44°56' .
We need to find the distance AC , which is the distance from observatory A to target C .
First, convert the angles from degrees, minutes, and seconds to decimal degrees for easier calculations.
1. Angle BAC = 52°42' :
- 42' = 4260 = 0.7° .
- So, BAC = 52 + 0.7 = 52.7° .
2. Angle ABC = 44°56' :
- 56' = 5660 ≈ 0.9333° .
- So, ABC = 44 + 0.9333 = 44.9333° .
In any triangle, the sum of the angles is 180° . Therefore:
The Law of Sines states:
We are solving for AC , so rearrange the formula:
Substitute the known values:
Using a calculator:
- sin(44.9333°) ≈ 0.7068 .
- sin(82.3667°) ≈ 0.9911 .
Now substitute these values into the equation:
The distance of the target C from observatory A is approximately:
Step 1:
Understand the Given Information- The distance between observatories A and B is AB = 296 m .
- Angle BAC = 52°42' .
- Angle ABC = 44°56' .
We need to find the distance AC , which is the distance from observatory A to target C .
Step 2:
Convert Angles to Decimal DegreesFirst, convert the angles from degrees, minutes, and seconds to decimal degrees for easier calculations.
1. Angle BAC = 52°42' :
- 42' = 4260 = 0.7° .
- So, BAC = 52 + 0.7 = 52.7° .
2. Angle ABC = 44°56' :
- 56' = 5660 ≈ 0.9333° .
- So, ABC = 44 + 0.9333 = 44.9333° .
Step 3:
Find Angle ACBIn any triangle, the sum of the angles is 180° . Therefore:
ACB = 180° - BAC - ABC
ACB = 180° - 52.7° - 44.9333°
ACB = 82.3667°
Step 4:
Apply the Law of SinesThe Law of Sines states:
ACsin(ABC) = ABsin(ACB)
We are solving for AC , so rearrange the formula:
AC = AB · sin(ABC)sin(ACB)
Substitute the known values:
AC = 296 · sin(44.9333°)sin(82.3667°)
Step 5:
Calculate the Sine ValuesUsing a calculator:
- sin(44.9333°) ≈ 0.7068 .
- sin(82.3667°) ≈ 0.9911 .
Now substitute these values into the equation:
AC = 296 · 0.70680.9911
AC ≈ 209.21280.9911
AC ≈ 211.1 m
Final Answer:
The distance of the target C from observatory A is approximately:
211.1 m
Tips for related online calculators
Do you want to convert time units like minutes to seconds?
Cosine rule uses trigonometric SAS triangle calculator.
See also our trigonometric triangle calculator.
Try conversion angle units angle degrees, minutes, seconds, radians, grads.
Cosine rule uses trigonometric SAS triangle calculator.
See also our trigonometric triangle calculator.
Try conversion angle units angle degrees, minutes, seconds, radians, grads.
You need to know the following knowledge to solve this word math problem:
planimetricsgoniometry and trigonometryUnits of physical quantitiesGrade of the word problem
We encourage you to watch this tutorial video on this math problem: video1
Related math problems and questions:
- Two artillery
Objective C we observe from two artillery observatories, A and B, which are 975 m apart. The size of the BAC angle is 63°, and the size of ABC is 48°. Calculate the distance of points A and C.
- Aircraft
From the aircraft flying at an altitude of 500m, they observed places A and B (at the same altitude) in the direction of flight at depth angles alpha = 48° and beta = 35°. What is the distance between places A and B?
- Bisectors
As shown, in △ ABC, ∠C = 90°, AD bisects ∠BAC, DE⊥AB to E, BE = 2, BC = 6. Find the perimeter of triangle △ BDE.
- Gamma angle
Find the magnitude of the gamma angle in triangle ABC if: α = 38° 56' and β = 47° 54'.
- Inaccessible 82710
Determine the distance between two inaccessible places K, L, if the angles KAL=62°10", LAB=41°23", KBL=66°34", and LBA were measured from points A, B, which are 870 m apart = 34°52". Thank you.
- Measurements 8129
The plane flies at an altitude of 22.5 km to the observatory. At the time of the first measurement, it was seen at an elevation angle of 28° and during the second measurement at an elevation angle of 50°. Calculate the distance it flies between these two
- As shown
As shown, in △ ABC, ∠C = 90°, AD bisects ∠BAC, DE⊥AB to E, BE = 2, BC = 6, then the perimeter of △ BDE