گام به گام تمرین صفحه 34 درس مثلثات ریاضی (1)

1)

\(\tan \,\theta  = \frac{h}{3} = \frac{{1/5}}{{0/5}} = 3 \Rightarrow h = 9\)

9 متر

2)

\(\begin{array}{l}\left. \begin{array}{l}B = {B_1} + {B_2} = {120^ \circ } \Rightarrow {B_1} = {B_2} = {60^ \circ }\\BH = \frac{1}{2}AB = \frac{3}{2}\\\tan \,{B_1} = \frac{{OH}}{{BH}} \Rightarrow OH = BH \times \tan \,{60^ \circ } = \frac{3}{2} \times \sqrt 3  = \frac{{3\sqrt 3 }}{2}\\{S_{A\mathop B\limits^\Delta  C}} = \frac{1}{2} \times AB \times OH = \frac{1}{2} \times 3 \times \frac{{3\sqrt 3 }}{2} = \frac{{9\sqrt 3 }}{4}\end{array} \right\}\\\\ \Rightarrow S = 6 \times {S_{A\mathop B\limits^\Delta  C}} = \frac{{27\sqrt 3 }}{2}\end{array}\)

3)

\(\tan \;O = \frac{{AH}}{{OA}} \Rightarrow OA = \frac{{AH}}{{\tan \,{{13}^ \circ }}} \simeq \frac{2}{{0/23}} \simeq 8/695\;Km\)

4)

\(\begin{array}{l}{S_{A\mathop B\limits^\Delta  C}} = \frac{1}{2} \times AH \times BC\\\sin \,B = \frac{{AH}}{{AB}} \Rightarrow AH = AB \times \sin \,{75^ \circ } = 3 \times 0/96 = 2/88\\{S_{A\mathop B\limits^\Delta  C}} = \frac{1}{2} \times 2/88 \times 5 = 7/2\end{array}\)

5)

\(\begin{array}{l}{S_{A\mathop B\limits^\Delta  C}} = \frac{1}{2} \times BH \times AC\\\sin \,A = \frac{{BH}}{{AB}} \Rightarrow BH = AB \times \sin \,{30^ \circ } = 3 \times \frac{1}{2} = \frac{3}{2}\\\cos \,A = \cos \,C = \frac{{AH}}{{AB}} = \frac{{HC}}{{BC}} \Rightarrow AH = HC = 3 \times \cos \,{30^ \circ } = \frac{{3\sqrt 3 }}{2}\\AC = AH + HC = 3\sqrt 3 \\{S_{A\mathop B\limits^\Delta  C}} = \frac{1}{2} \times \frac{3}{2} \times 3\sqrt 3  = \frac{{9\sqrt 3 }}{4}\end{array}\)