Question

Triunghiul ABC AB=4, AC= radical din 7, BC=radical din 3 sa se afle cos B? Trigoonometrie cl 9

Answer 1

a = √3  ; b = √7    , c = 4 
cosB = [ a² + c² - b² ] / 2·a·c = [ (√3)²  + 4² - (√7)² ]  / 2 ·√3·4 = [ 3 + 16  -7 ] /8√3 = 
 = 12 / 8 ·√3 = 3 /2√3 = 3√3 / 2·√3·√3 =  3√3 / 6  = √3 / 2 
                                   ⇒ mas<B = 30⁰

Answer 2

$cos~B= \frac{a^2+c^2-b^2}{2ac} = \frac{(\sqrt{3})^2+4^2- ( \sqrt{7})^2}{2 \cdot \sqrt{3} \cdot 4}= \frac{3+16-7}{8 \sqrt{3}}= \frac{12}{8 \sqrt{3}}= \frac{3}{2 \sqrt{3}}= \frac{3 \sqrt{3}}{2 \cdot3}= \\ \\ = \frac{\sqrt{3}}{2}. \\ \\ Mai~mult:~cos~B= \frac{\sqrt{3}}{2} \Rightarrow m(\ \textless \ B)=30 \textdegree. \\ \\ Am~notat:~a=BC;~b=AC;~c=AB.$