Find Exact Value of sin x in Right Triangle ABC with Given Hypotenuse and Angle

Question

Accepted Answer

sin x = 3/13

Answer

Step 1: In right triangle ABD, use the tangent function to find AB. Since ∠DAB = π/6 and DB = 4, tan(π/6) = DB/AB. We know tan(π/6) = 1/√3, so 1/√3 = 4/AB, which gives AB = 4√3. Step 2: In right triangle ABC, use the Pythagorean theorem to find BC. Since AC = 13 and AB = 4√3, BC = √(AC² - AB²) = √(13² - (4√3)²) = √(169 - 48) = √121 = 11. Then calculate DC = BC - DB = 11 - 4 = 7. Step 3: Calculate ∠BAC. In right triangle ABC, sin(∠BAC) = BC/AC = 11/13, cos(∠BAC) = AB/AC = 4√3/13. Note that ∠BAC = x + π/6. Step 4: Use the sine addition formula: sin(x + π/6) = sin x cos(π/6) + cos x sin(π/6) = 11/13. Substitute cos(π/6) = √3/2 and sin(π/6) = 1/2, so (√3/2)sin x + (1/2)cos x = 11/13. Multiply both sides by 2: √3 sin x + cos x = 22/13, so cos x = 22/13 - √3 sin x. Step 5: Use the Pythagorean identity sin²x + cos²x = 1. Substitute cos x = 22/13 - √3 sin x into the identity: sin²x + (22/13 - √3 sin x)² = 1. Expand the equation: sin²x + (484/169 - (44√3/13)sin x + 3sin²x) = 1. Combine like terms: 4sin²x - (44√3/13)sin x + 484/169 - 1 = 0, which simplifies to 4sin²x - (44√3/13)sin x + 315/169 = 0. Multiply through by 169 to eliminate denominators: 676sin²x - 748√3 sin x + 315 = 0. Step 6: Solve the quadratic equation for sin x using the quadratic formula sin x = [748√3 ± √((748√3)² - 4*676*315)]/(2*676). Calculate the discriminant: (748√3)² - 4*676*315 = 748²*3 - 4*676*315 = (4*187)²*3 - 4*676*315 = 4*(187²*3 - 676*315) = 4*(104367 - 213060) = 4*(-108693). This indicates an error in the addition formula method, so switch to the Sine Rule method. Step 7: Apply the Sine Rule to triangle ADC and triangle ABD. First, in triangle ABD, AD = DB / sin(π/6) = 4 / (1/2) = 8. In triangle ADC, ∠ACD = arcsin(AB/AC) = arcsin(4√3/13), ∠ADC = π - ∠ADB = π - (π - π/6 - π/2) = 2π/3. Then by the Sine Rule in triangle ADC: AD/sin(∠ACD) = AC/sin(∠ADC). Substitute values: 8/sin(∠ACD) = 13/sin(2π/3). We know sin(2π/3) = √3/2, so sin(∠ACD) = (8*(√3/2))/13 = 4√3/13, which matches the earlier value. Then use the Sine Rule directly for angle x: DC/sin x = AC/sin(∠ADC). So 7/sin x = 13/(√3/2), which gives sin x = (7*(√3/2))/13 = 7√3/26. This is incorrect, so go back to the right triangle angle relationship. Step 8: Correct method: In right triangle ABC, sin(∠BAC) = 11/13, ∠BAC = x + π/6. Use sin(x + π/6) = 11/13, and cos(x + π/6) = 4√3/13. Use the identity sin x = sin[(x + π/6) - π/6] = sin(x + π/6)cos(π/6) - cos(x + π/6)sin(π/6). Substitute values: sin x = (11/13)(√3/2) - (4√3/13)(1/2) = (11√3 - 4√3)/26 = 7√3/26. Wait, no, correct AB calculation: tan(π/6) = DB/AB => AB = DB / tan(π/6) = 4/(1/√3) = 4√3, correct. BC = 11, correct. Then in triangle ADC, use the Sine Rule: AD = 8, DC =7, AC=13, ∠ADC = π - ∠ADB. ∠ADB = π - π/6 - π/2 = π/3, so ∠ADC = 2π/3. Then sin x / DC = sin(∠ADC)/AC => sin x = (7 * sin(2π/3))/13 = (7*(√3/2))/13 =7√3/26. But wait, another way: use cos(∠BAC)=AB/AC=4√3/13, ∠BAC=x+π/6, so cos(x+π/6)=4√3/13. Expand cos(x+π/6)=cosx cos(π/6)-sinx sin(π/6)=4√3/13. And sin(x+π/6)=11/13. Solve the system: (√3/2)cosx - (1/2)sinx=4√3/13 and (√3/2)sinx + (1/2)cosx=11/13. Multiply the first equation by √3: (3/2)cosx - (√3/2)sinx=12/13. Add to the second equation: 2cosx=12/13 +11/13=23/13 => cosx=23/26. Then substitute into (√3/2)sinx + (1/2)(23/26)=11/13 => (√3/2)sinx=11/13 -23/52= (44-23)/52=21/52 => sinx=(21/52)*(2/√3)=21/(26√3)=7√3/26. This is the correct exact value.