1. W trójkącie prostokątnym naprzeciw kąta ostrego α leży przyprostokątna długości a. Oblicz długość pozostałych boków trójkąta jeśli a = 5, tg = 2
2. Oblicz: sin 120 * tg 300 / (cos 210 * tg 150)
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tgα = a/b
2 = 5/b
b = 2,5
5²+2,5²=c²
c² = 25+6,25
c = 25√0,05
2.![\frac{sin 120*tg300}{cos210*tg150}\\sin120 = sin(90+30) = cos30 = \frac{\sqrt{3}}{2}\\tg300 = tg(270+30) = ctg30 = \sqrt{3}\\cos210 = cos(270-60)=sin60 = \frac{\sqrt{3}}{2}\\tg150 = tg(90+60) = ctg60 = \frac{\sqrt{3}}{3}\\\frac{ \frac{\sqrt{3}}{2}* \sqrt{3}}{\frac{\sqrt{3}}{2}*\frac{\sqrt{3}}{3}} = \frac{\frac{3}{2}}{\frac{3}{6}} = \frac{3}{2}*\frac{2}{1} = 3 \frac{sin 120*tg300}{cos210*tg150}\\sin120 = sin(90+30) = cos30 = \frac{\sqrt{3}}{2}\\tg300 = tg(270+30) = ctg30 = \sqrt{3}\\cos210 = cos(270-60)=sin60 = \frac{\sqrt{3}}{2}\\tg150 = tg(90+60) = ctg60 = \frac{\sqrt{3}}{3}\\\frac{ \frac{\sqrt{3}}{2}* \sqrt{3}}{\frac{\sqrt{3}}{2}*\frac{\sqrt{3}}{3}} = \frac{\frac{3}{2}}{\frac{3}{6}} = \frac{3}{2}*\frac{2}{1} = 3](https://tex.z-dn.net/?f=%5Cfrac%7Bsin+120%2Atg300%7D%7Bcos210%2Atg150%7D%5C%5Csin120+%3D+sin%2890%2B30%29+%3D+cos30+%3D+%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%5C%5Ctg300+%3D+tg%28270%2B30%29+%3D+ctg30+%3D+%5Csqrt%7B3%7D%5C%5Ccos210+%3D+cos%28270-60%29%3Dsin60+%3D+%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%5C%5Ctg150+%3D+tg%2890%2B60%29+%3D+ctg60+%3D+%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B3%7D%5C%5C%5Cfrac%7B+%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%2A+%5Csqrt%7B3%7D%7D%7B%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%2A%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B3%7D%7D+%3D+%5Cfrac%7B%5Cfrac%7B3%7D%7B2%7D%7D%7B%5Cfrac%7B3%7D%7B6%7D%7D+%3D+%5Cfrac%7B3%7D%7B2%7D%2A%5Cfrac%7B2%7D%7B1%7D+%3D+3)