Parallelogram 82695

Given is the parallelogram KLMN, in which we know the side sizes/KL/ = a = 84.5 cm, /KN/ = 47.8 cm, and the angle size at the vertex K 56°40'. Calculate the size of the diagonals.

Correct answer:

u₁ =  117.7461 cm
u₂ =  70.6119 cm

Step-by-step explanation:

$$\begin{gathered} a=84.5\text{cm} \\ b=47.8\text{cm} \\ K=56°40^{\prime }=56°+\frac{40}{60}°=56.6667°\doteq 56.6667 \\ \sin K=h:b \\ h=b\cdot \sin K=b\cdot \sin 56.666666666667°=47.8\cdot \sin 56.666666666667°=47.8\cdot 0.835488=39.93632\text{cm} \\ x=b\cdot \cos K=b\cdot \cos 56.666666666667°=47.8\cdot \cos 56.666666666667°=47.8\cdot 0.549509=26.26653\text{cm} \\ {u}_1=\sqrt{h^2+(a+x{)}^2}=\sqrt{39.9363^2+(84.5+26.2665{)}^2}=117.7461\text{cm} \end{gathered}$$

$$\begin{gathered} L=180-K=180-56.6667=\frac{370}{3}\doteq 123.3333{}^{\circ } \\ \sin L=g:a \\ g=a\cdot \sin L=a\cdot \sin 123.33333333333°=84.5\cdot \sin 123.33333333333°=84.5\cdot 0.835488=70.59872\text{cm} \\ y=a\cdot \cos L=a\cdot \cos 123.33333333333°=84.5\cdot \cos 123.33333333333°=84.5\cdot (-0.549509)=-46.43351\text{cm} \\ {u}_2=\sqrt{g^2+(b+y{)}^2}=\sqrt{70.5987^2+(47.8+(-46.4335){)}^2}=70.6119\text{cm} \end{gathered}$$

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