Mathematics in Ancient and Medieval India c6.s9.50

c6.s9.50

In △ ADC, $AC² = c² + d² - 2cd (R Cos D)/R$ $= c² + d² - 2cd (R Cos(180⁰--B))/R$ $= c² + d² + 2cd (R Cos B)/R$ $∴ a² + b² - (2ab R Cos B)/R = c² + d² + 2cd (R Cos B)/R$ $or (R Cos B)/R = (a² + b² - c² - d²)/(2(ab + cd)) ··· ··· ··· (4)$ Now (R Sin B)² = R² - R² Cos²B = R²1 - ((R Cos B)/R)² $= R²[1 - (a² + b² - c² - d²)/(2(ab + cd))²]$ $= R²(2(ab + cd)² - (a² + b² - c² - d²)²)/(4(ab + cd)²)$ $= R²((a + b)² - (c - d)²(c + d)² - (a - b)²)/(4(ab + cd)²)$ $= R²((a + b + c--d)(a + b--c + d)(c + d + a--b)(c + d--a + b))/(4(ab + cd)²)$ $= R² (2(s - d). 2(s - c). 2(s - b). 2(s - a))/(4(ab + cd)²) ( when$ $2s = a + b + c + d.)$ $= R² (16.(s - a)(s - b) (s - c) (s - d))/(4(ab + cd)²)$ $∴ (R Sin B)/R = 2 √(((s - a)(s - b)(s - c) (s - d))/((ab + cd))) ··· ···(5)$ Now quad. ABCD $= +$