Answer to Question #77687 in Trigonometry for Sarai

Question

Two points, A and B, are separated by an obstacle. The straight line DAC is run making AC = 550 ft and AD = 1100 ft. The angle BCA is found to be 57° and angle BDA to be 47°. Find the distance AB.

Answer 1

Let us consider triangle DCB.
It’s clear that DC = AD + AC = 550 + 1100 = 1650 ft, and angle DBC = 180 – 57 − 47 = 76°
Using sine’s theorem we can find, for example, DB: DC/sin⁡〖76°〗 = DB/sin⁡〖57°〗 , hence DB = sin⁡〖57°〗/sin⁡〖76°〗 DC = 0.838671/0.970296 1650 =1426.17 ft
Then, using cosine’s theorem, we can find AB: AB² =DB² +DA² -2 ×DB× DA × cos⁡〖47°〗
Thus AB² = 1426.17² + 1100² − 2 × 1426.17 × 1100 × 0.681998 = 1104141.676048
And AB = 1050.78 ft