# Right

Determine angles of the right triangle with the hypotenuse c and legs a, b, if:

$3a+4b=4.9c$

Correct result:

α =  64.6 °
β =  25.4 °
γ =  90 °

#### Solution:

$\ \\ 3a +4b = 4.9c \ \\ \ \\ c=1 \ \\ 3a +4b = 4.9 \ \\ a^2+b^2 =1 \ \\ \ \\ a=\dfrac{ 4.9 - 4 b}{ 3} \ \\ \ \\ \dfrac{ (4.9 - 4 b)^2}{ 3^2} + b^2 =1 \ \\ (4.9 - 4 b)^2 + 3^2 b^2 = 3^2 \ \\ \ \\ \ \\ 25b^2 -39.2b +15.01 =0 \ \\ \ \\ p=25; q=-39.2; r=15.01 \ \\ D = q^2 - 4pr = 39.2^2 - 4\cdot 25 \cdot 15.01 = 35.64 \ \\ D>0 \ \\ \ \\ b_{1,2} = \dfrac{ -q \pm \sqrt{ D } }{ 2p } = \dfrac{ 39.2 \pm \sqrt{ 35.64 } }{ 50 } \ \\ b_{1,2} = 0.784 \pm 0.119398492453 \ \\ b_{1} = 0.903398492453 \ \\ b_{2} = 0.664601507547 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ 25 (b -0.903398492453) (b -0.664601507547) = 0 \ \\ \ \\ \ \\ \alpha = \arcsin b = 64.6 ^\circ$
$\beta =\mathrm{arccos}b=25.{4}^{\circ }$
$\gamma =9{0}^{\circ }$

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