Question

In triangle ABC, BC=3, AC=4, AB=5, then the value of sinA+sin2B+sin3C​

Answer 1

Answer:

This is your solution

Answer 2

Answer:

14/25

Step-by-step explanation:

Given

let BC a =3, AC b=4, AB c=5

Here angle c is 90degrees

remaining should be 90 degrees

as sum of the angles in triangle should be 180

$△= \sqrt{s(s−a)(s−b)(s−c)} =6</p><p></p><p>$

$\cos(b) = \frac{ {a}^{2} + {c}^{2} - {b}^{?} }{2ac}$

$= \frac{ {3}^{2} + {4}^{2} - {5}^{2} }{2 \times 3 \times 4}$

$= \frac{18}{30}$

$\sin(A) = \frac{2}{?}$

$sinA= \frac{2△}{bc} = \frac{5}{3}$

$\sin(b) = </p><p> \frac{△</p><p> \times 2}{ac} = \frac{4}{5}$

$sinC= \frac{2△}{ab} = 1$

Now sinA+sin2B+sin3C

$=sinA+2sinBcosB+3sinC - 4 {\sin }^{3} c$

$= \frac{14}{25}$

#SPJ3