Right

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

3a+4b=4.9c3a +4b = 4.9c


Correct result:

α =  64.6 °
β =  25.4 °
γ =  90 °

Solution:

 3a+4b=4.9c  c=1 3a+4b=4.9 a2+b2=1  a=4.9−4b3  (4.9−4b)232+b2=1 (4.9−4b)2+32b2=32   25b2−39.2b+15.01=0  p=25;q=−39.2;r=15.01 D=q2−4pr=39.22−4⋅25⋅15.01=35.64 D>0  b1,2=−q±D2p=39.2±35.6450 b1,2=0.784±0.11939849245279 b1=0.90339849245279 b2=0.66460150754721   Factored form of the equation:  25(b−0.90339849245279)(b−0.66460150754721)=0   α=arcsin⁡b=64.6∘ \ \\ 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 0.11939849245279 \ \\ b_{1} = 0.90339849245279 \ \\ b_{2} = 0.66460150754721 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ 25 (b -0.90339849245279) (b -0.66460150754721) = 0 \ \\ \ \\ \ \\ \alpha = \arcsin b = 64.6 ^\circ
β=arccos⁡b=25.4∘ \beta = \arccos b = 25.4 ^\circ
γ=90∘\gamma = 90 ^\circ

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