Geometry Problem: Finding Angle BAD / sin CAD in Triangle ABC
Introduction to the Geometry Problem
In triangle ABC, angles B and C are given as B π/3 (60°) and C π/4 (45°). Angle A is therefore 75°. Point D divides BC internally in the ratio 1:3. The problem is to find the ratio of Angle BAD to sin Angle CAD.
Step-by-Step Solution
Let BD x and DC 3x. Thus, BC 4x. We can use the Law of Sines in triangle ABC to find the side lengths:
c/sin 45 b/sin 60 a/sin 75
c/sin 45 4x/sin 75, thus c 4x * sin 45 / sin 75 2.92820323x
b 4x * sin 60 / sin 75 3.586301889x
Calculating AD
In triangle ABD:
AD^2 AB^2 BD^2 - 2AB*BD*cos(B)
AD^2 (2.92820323x)^2 (x)^2 - 2 * 2.92820323x * x * cos(60°)
AD^2 8.574374158x^2 - x^2 6.646170928x^2
AD 2.578016859x
Further Calculations
In triangle ACD:
AD^2 AC^2 CD^2 - 2AC*CD*cos(C)
AD^2 (3.586301889x)^2 (3x)^2 - 2 * 3.586301889x * 3x * cos(45°)
AD^2 12.86156124x^2 9x^2 - 15.21539031x^2 6.646170928x^2
AD 2.578016859x
In triangle ABD, using the Law of Sines:
c/sin ADB x/sin BAD 2.578016859x/sin 60
sin ADB 2.92820323x * sin 60 / 2.578016859x 0.983662451
ADB arcsin(0.983662451) 79.62891666°
In triangle ACD:
b/sin ADC 3x/sin CAD 2.578016859x/sin 45
sin ADC 3.586301889x * sin 45 / 2.578016859x 0.983662452
ADC arcsin(0.983662452) 100.3710834° (Corrected Check)
Therefore, BAD ADC - 60° 180° - 79.62891666° - 60° 40.37108334°
Final Ratio
Angle CAD 75° - BAD 75° - 40.37108334° 34.62891666°
Thus, the ratio: BAD / sin CAD 40.37108334 / sin(34.62891666) 71.04344336
Final Answer
Therefore, the final answer is 71.04344336.
Keywords
The keywords for this article are: Triangle, Geometry, Trigonometry, Angles, Ratio