Here's the solution,
figure 1.
by using trigonometry,
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[tex] \sin(60) = \dfrac{5 \sqrt{3} }{y} [/tex]
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[tex] \dfrac{ \sqrt{3} }{2} = \dfrac{5 \sqrt{3} }{y} [/tex]
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[tex] \dfrac{1}{y} = \dfrac{ \sqrt{3} }{2 \times 5 \sqrt{3} } [/tex]
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[tex] \dfrac{1}{y} = \dfrac{1}{10} [/tex]
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[tex]y = 10[/tex]
And,
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[tex] \cos(60) = \dfrac{w}{y} [/tex]
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[tex] \dfrac{1}{2} = \dfrac{w}{10} [/tex]
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[tex]w = 5[/tex]
So, values are :
figure 2.
Since, it's an isosceles triangle, w = y,
So, by pythagoras theorem :
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[tex] {w}^{2} + {y}^{2} = (7 \sqrt{2} ) {}^{2} [/tex]
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[tex] {w}^{2} + {w}^{2} = 98[/tex]
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[tex]2 {w}^{2} = 98[/tex]
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[tex] {w}^{2} = 49[/tex]
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[tex]w = \sqrt{49} [/tex]
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[tex]w = 7[/tex]
and we know, w = y, so their values will be equal to 7 units.