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Sprendimas:

∠SBA = 30°

tg ∠SBA = $\frac{SA}{AB}$

$tg(30)$ = $\frac{h}{AB}$

$\frac{\sqrt {3}}{3}$ = $\frac{h}{AB}$

$AB$ = $\frac{3\cdot h}{\sqrt {3}}$ = $\frac{3\cdot h\cdot \sqrt {3}}{3}$ = $h\cdot \sqrt {3}$

Atsakymas: AB = $h\cdot \sqrt {3}$

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Sprendimas:

∠SCA = 45°, △SCA - status, lygiašonis, todėl AC = AS = h;

Žinome visas tris trikampio △ABC kraštines ir kampą ∠BAC, pritaikysime kosinusų teoremą:

$BC^{2}$ = $AC^{2}+AB^{2}-2\cdot AC\cdot AB\cdot cos(BAC)$

$105^{2}$ = $h^{2}+(h\cdot \sqrt {3})^{2}-2\cdot h\cdot h\cdot \sqrt {3}\cdot cos(150)$

$105^{2}$ = $h^{2}+3\cdot h^{2}-2\cdot h\cdot h\cdot \sqrt {3}\cdot cos(150)$

$105^{2}$ = $h^{2}+3\cdot h^{2}+\frac{2\cdot h\cdot h\cdot \sqrt {3}\cdot \sqrt {3}}{2}$

$105^{2}$ = $h^{2}+3\cdot h^{2}+\frac{2\cdot h\cdot h\cdot 3}{2}$

$105^{2}$ = $h^{2}+3\cdot h^{2}+3\cdot h^{2}$

$105^{2}$ = $7\cdot h^{2}$

$(15\cdot 7)^{2}$ = $7\cdot h^{2}$

$h^{2}$ = $\frac{15^{2}\cdot 7^{2}}{7}$

$h^{2}$ = $15^{2}\cdot 7$

$h$ = $\sqrt {15^{2}\cdot 7}$

$h$ = $15\cdot \sqrt {7}$

Atsakymas: $15\cdot \sqrt {7}$

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