Question

please answer this question with whole method

Answer 1

good question...

see, ABC IS a right angled triangle
apply Pythagoras theorem to find the value of x.

$(5 - x) {}^{2} + {x}^{2} = { (\sqrt{13}) }^{2} \\ 25 + 2 {x}^{2} - 10x = 13 \\ 2 {x}^{2} - 10x + 12 = 0 \\ {x}^{2} - 5x + 6 = 0 \\ {x}^{2} - 2x - 3x + 6 = 0 \\ x(x - 2) - 3(x - 2) = 0 \\ (x - 3)(x - 2) = 0 \\ x = 3 \: or \: 2 \:$
if you will put x=3
sides will be 2,3,√13

and if you will put x=2
sides will be 2,3,√13

since sides are same... no matter what value you take

then.,
if u take x=2
$\sin( \alpha ) + \sin( \beta ) \\ = \frac{2}{ \sqrt{13} } + \frac{3}{ \sqrt{13} } \\ = \frac{5}{ \sqrt{13} }$


mark as brainliest if helped

Answer 2

sin a +sin B
sin a + sin b