Question

triunghi ABC , AB=10 cm , BC=24cm , AC=26cm . Calculati : b) sin  (<ACB)

Answer 1

  daca ΔABC dreptunghic in A ⇒ sin <ACB = AB/BC = 10/24=5/12

Answer 2

$A \ tr ABC= \sqrt{p(p-a)(p-b)(p-c)} \\ p= \frac{a+b+c}{2} = \frac{10+26+24}{2}= \frac{60}{2}=30 \\ A = \sqrt{30(30-10)(30-24)(30-26)} = \sqrt{3*10 * 5 * 20 *6*4} \\ A= \sqrt{3*2*5*4*5 *2*3 *4} = 3*2*5*4= 120$

$A\ trABC= \frac{b*h}{2}$

fie [AD _|_ BC
[AD]-h

$A \ trABC=120 \\ A\ trABC= \frac{AD*BC}{2} = \frac{AD*24}{2} \\ \frac{AD*24}{2}=120 =>AD= \frac{2*120}{24}= 10$

ΔADC, mas<D=90

sin<ACD= AD/AC= 10/26 = 5/13

punctul D ≡ B  (se confunda)  => sin<ACB= 5/13