Răspuns :

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

[tex]A\ trABC= \frac{b*h}{2} [/tex]

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

[tex]A \ trABC=120 \\ A\ trABC= \frac{AD*BC}{2} = \frac{AD*24}{2} \\ \frac{AD*24}{2}=120 =>AD= \frac{2*120}{24}= 10[/tex]

ΔADC, mas<D=90

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

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