Question

in triangle abc right angled at b ab=5 cm sin c = 1/2​

Answer 1

Answer:

$\frac{25\sqrt{3} }{2}$

Step-by-step explanation:

• B^ = 90

• AB = 5cm

• sinC = 1/2

sinC = $\frac{AB}{AC}$

$\frac{1}{2} =\frac{5}{AC}$

• $AC= 5 . 2 = 10$

$AB^{2} + BC^{2} = AC^{2}$

$5^{2} +BC^{2} =10^{2}$

$BC^{2} =75$

• $BC=5\sqrt{3}$

• S= $\frac{1}{2} AB.BC=\frac{1}{2} 5.5\sqrt{3}$=$\frac{25\sqrt{3} }{2}$

Answer 2

Answer:

In ΔABC ,

∠B=90

∘

AB=5cm

sinC=

2

1

AC=?

Now,

sinC=

AC

AB

2

1

=

AC

5

AC=5×2=10cm