Bài 33 trang 66 SGK Hình học 10 nâng cao

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3.4 trên 5 phiếu

Giải tam giác ABC, biết

Bài 33. Giải tam giác $$ABC$$, biết

a) $$c = 14,\,\widehat A = {60^0},\,\widehat B = {40^0}$$;

b) $$b = 4,5,\,\widehat A = {30^0},\,\widehat C = {75^0}$$;

c) $$c = 35,\,\widehat A = {40^0},\,\widehat C = {120^0}$$;

d) $$a = 137,5;\;\widehat B = {83^0},\,\widehat C = {57^0}$$.

Hướng dẫn trả lời

a)  Ta có $$\widehat C = {180^0} - {60^0} - {40^0} = {80^0}$$

Áp dụng định lí sin :

\eqalign{ & \,\,\,\,\,\,{a \over {\sin A}} = {b \over {\sin B}} = {c \over {\sin C}} = {{14} \over {{\mathop{\rm s}\nolimits} {\rm{in8}}{{\rm{0}}^0}}}\,\,\,\, \Rightarrow \,\,a = {{14} \over {{\mathop{\rm s}\nolimits} {\rm{in8}}{{\rm{0}}^0}}}.\sin {60^0} \approx 12,3 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b = {{14} \over {{\mathop{\rm s}\nolimits} {\rm{in8}}{{\rm{0}}^0}}}.\sin {40^0} \approx 9,1 \cr}

b) Ta có $$\widehat B = {180^0} - {30^0} - {75^0} = {75^0}$$

Áp dụng định lí sin

\eqalign{ & \,\,\,\,\,\,{a \over {\sin A}} = {b \over {\sin B}} = {c \over {\sin C}} = {{4,5} \over {{\mathop{\rm s}\nolimits} {\rm{in7}}{{\rm{5}}^0}}}\,\,\, \Rightarrow \,\,a = {{4,5} \over {{\mathop{\rm s}\nolimits} {\rm{in7}}{{\rm{5}}^0}}}.\sin {30^0} \approx 2,3 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c = {{4,5} \over {{\mathop{\rm s}\nolimits} {\rm{in7}}{{\rm{5}}^0}}}.\sin {75^0} = 4,5 \cr}

c)  Ta có $$\widehat B = {180^0} - {120^0} - {40^0} = {20^0}$$

Áp dụng định lí sin :

\eqalign{ & \,\,\,\,\,\,{a \over {\sin A}} = {b \over {\sin B}} = {c \over {\sin C}} = {{35} \over {{\mathop{\rm s}\nolimits} {\rm{in12}}{{\rm{0}}^0}}}\,\,\,\,\, \Rightarrow \,\,a = {{35} \over {{\mathop{\rm s}\nolimits} {\rm{in12}}{{\rm{0}}^0}}}.\sin {40^0} \approx 26 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b = {{35} \over {{\mathop{\rm s}\nolimits} {\rm{in12}}{{\rm{0}}^0}}}.\sin {20^0} \approx 13,8 \cr}

d)  Ta có $$\widehat A = {180^0} - {83^0} - {57^0} = {40^0}$$

Áp dụng định lí sin :

\eqalign{ & \,\,\,\,\,\,{a \over {\sin A}} = {b \over {\sin B}} = {c \over {\sin C}} = {{137,5} \over {{\mathop{\rm s}\nolimits} {\rm{in4}}{{\rm{0}}^0}}}\,\,\,\, \Rightarrow \,\,b = {{137,5} \over {{\mathop{\rm s}\nolimits} {\rm{in4}}{{\rm{0}}^0}}}.\sin {83^0} \approx 212,3 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c = {{137,5} \over {{\mathop{\rm s}\nolimits} {\rm{in4}}{{\rm{0}}^0}}}.\sin {57^0} \approx 179,4 \cr}

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