Bài 9 trang 108 Tài liệu dạy – học Toán 9 tập 1

Đề bài

Giải tam giác ABC vuông tại A, biết rằng

a) $b = 16cm,\widehat C = {30^o}$;

b) $c = 24cm,\widehat C = {60^o}$;

c) $a = 20cm,\widehat C = {45^o}$;

d) $a = 10cm;\widehat B = {38^o}$;

e) $c = 21cm,b = 18cm$.

Lời giải chi tiết

a) $b = 16cm,\widehat C = {30^o}$;

$\begin{array}{l}c = b.\tan \widehat C = 16.\tan {30^o} \approx 9,24\,(cm)\\b = a.\cos \widehat C \Rightarrow a = \dfrac{b}{{\cos \widehat C}} = \dfrac{{16}}{{\cos {{30}^o}}} \approx 18,48\,\,(cm)\\\widehat B = {90^o} – \widehat C = {90^o} – {30^o} = {60^o}\end{array}$

b) $c = 24cm,\widehat C = {60^o}$;

$\begin{array}{l}b = c.\cot \widehat C = 24.\cot {60^o} \approx 13,86\,\,(cm)\\c = a.\sin \widehat C \Rightarrow a = \dfrac{c}{{\sin \widehat C}} = \dfrac{{24}}{{\sin {{60}^o}}} \approx 27,71\,(cm)\\\widehat B = {90^o} – \widehat C = {90^o} – {60^o} = {30^o}\end{array}$

c) $a = 20cm,\widehat C = {45^o}$;

$\begin{array}{l}c = a.\sin \widehat C = 20.\sin {45^o} \approx 14,14\,\,(cm)\\b = a.\cos \widehat C = 20.\cos {45^o} \approx 14,14\,\,(cm)\\\widehat B = {90^o} – \widehat C = {90^o} – {45^o} = {45^o}\end{array}$

d) $a = 10cm;\widehat B = {38^o}$;

$\begin{array}{l}b = a.\sin \widehat B = 10.\sin {38^o} \approx 6,16\,\,(cm)\\c = a.\cos \widehat B = 10.\cos {38^o} \approx 7,88\,\,(cm)\\\widehat C = {90^o} – \widehat B = {90^o} – {38^o} = {52^o}\end{array}$

e) $c = 21cm,b = 18cm$.

Áp dụng định lý Pythagore: $a = \sqrt {{b^2} + {c^2}}  = \sqrt {{{18}^2} + {{21}^2}}  \approx 27,66\,\,(cm)$

$\begin{array}{l}\tan \widehat B = \dfrac{b}{c} = \dfrac{6}{7} \Rightarrow \widehat B \approx {40^o}36’\\\widehat C = {90^o} – \widehat B \approx {49^o}24’\end{array}$

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