\(\triangle\)ABC is a right-angle triangle at B and tan A = \(\frac{3}{4}\), then sin A + sin B + sin C will be equal to: 

Given:

ΔABC is a right-angle triangle at B and tan A = \(\frac{3}{4}\)

Concept used:

1. tan θ = Height ÷ Base

2. sin θ = Height ÷ Hypotenuse

3. Pythagoras theorem states that “In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides“.

Calculation:

tan A = \(\frac{3}{4}\)

⇒ \(\frac {BC}{AB}\) = \(\frac{3}{4}\)

⇒ BC : AB = 3 : 4

Let the measure of BC and AB be 3k and 4k respectively.

​According to the concept,

AC² = \(\sqrt { (3k)^2 + (4k)^2 }\) = 5k

Now, sin A + sin B + sin C

⇒ \(\frac {BC}{AC} + sin\ 90^\circ + \frac {AB}{AC}\)

⇒ \(\frac {BC + AB}{AC} + 1\)

⇒ \(\frac {3k + 4k}{5k} + 1\)

⇒ \(\frac {7}{5} + 1\)

⇒ \(\frac {12}{5}\) = \(2\frac{2}{5}\)

∴ The value of sin A + sin B + sin C is \(2\frac{2}{5}\).