23 uždavinys

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