Area of triangle

1) b² = a² + c² - 2ac cosB5² = a² + 10.5² - 2a 10.5 cos25°a² - 21cos25° a + 85.25 = 0a = { 21cos25° ± √ [(21cos25°)² - 4(85.25)] } / 2a = 7.212177 or a = 11.820286ΔABC = (1/2) ac sinB = (1/2) (7.212177) (10.5) sin25° or (1/2) (11.820286) (10.5) sin25° = 16.002 cm² or 26.226211 cm²
Alternatively :sinC / 10.5 = sin25° / 5
sinC = 0.887498
C = 62.560555° or 117.439445°A = 180 - 25 - 62.560555 = 92.439445°
or
A = 180 - 25 - 117.439445° = 37.560555°ΔABC = (1/2) (10.5) (5) sin 92.439445° = 26.226211 cm²
or
ΔABC = (1/2) (10.5) (5) sin 37.560555° = 16.002 cm²
2a) Let a , b , c be 4k , 5k , 7k ,a + b + c = 4k + 5k + 7k = 64k = 4a = 4*4 = 16
b = 5*4 = 20
c = 7*4 = 28ΔABC
= √[s(s - a)(s - b)(s - c)]
= √[(64/2)(64/2 - 16)(64/2 - 20)(64/2 - 28)]
= 156.767 cm²
Alternatively :cosC = (16² + 20² - 28²) / (2*16*20) = - 0.2
C = 101.536959
sinC = 0.979796ΔABC
= (1/2) ab sinC
= (1/2) (16) (20) (0.979796)
= 156.767 cm²
b)Let h be the height from C to AB :(1/2) AB h = 156.767 (1/2) 28 h = 156.767 h = 11.198 cm