Cho tam giác \(ABC\) có \(c = 4,b = 7,\widehat A = {60^ \circ }\). Tính chiều cao \({h_a}\) của tam giác ABC (làm tròn đến hàng đơn vị)?
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Ta có: \(c = 4,b = 7,\widehat A = {60^ \circ }\)
Áp dụng định lí cosin trong tam giác ABC ta có:
\({a^2} = {b^2} + {c^2} - 2bc\cos A\)
\(\begin{array}{l} \Leftrightarrow {a^2} = {7^2} + {4^2} - 2.7.4\cos {60^ \circ } = 37\\ \Rightarrow a = \sqrt {37} \end{array}\)
Lại có: \(S = \frac{1}{2}b.c\sin A = \frac{1}{2}a.{h_a} \Rightarrow {h_a} = \frac{{b.c\sin A}}{a} = \frac{{7.4.\sin {{60}^ \circ }}}{{\sqrt {37} }} \approx 4\)
Vậy độ dài đường cao \({h_a}\) là khoảng 4.
Chọn C.
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