cho tam giác ABC có góc A=60 độ, a=10, r=5 căn 3/3. tính R,b,c 17/07/2021 Bởi Natalia cho tam giác ABC có góc A=60 độ, a=10, r=5 căn 3/3. tính R,b,c
Đáp án: \(R = \dfrac{{10\sqrt 3 }}{3},b = 10,c = 10\) Giải thích các bước giải: Áp dụng định lí sin: \(\dfrac{a}{{\sin A}} = 2R \Leftrightarrow R = \dfrac{a}{{2\sin A}} = \dfrac{{10}}{{2\sin {{60}^0}}} = \dfrac{{10\sqrt 3 }}{3}\) Ta có: \(\begin{array}{l}S = \dfrac{{abc}}{{4R}} = pr \Leftrightarrow \dfrac{{10bc}}{{4.\dfrac{{10\sqrt 3 }}{3}}} = \dfrac{{10 + b + c}}{2}.\dfrac{{5\sqrt 3 }}{3}\\ \Leftrightarrow 60bc = \dfrac{{40\sqrt 3 }}{3}.5\sqrt 3 \left( {10 + b + c} \right)\\ \Leftrightarrow 60bc = 200\left( {10 + b + c} \right)\\ \Leftrightarrow 3bc = 10\left( {10 + b + c} \right)\,\,\left( 1 \right)\end{array}\) Áp dụng định lý cos: \(\begin{array}{l}{a^2} = {b^2} + {c^2} – 2bc\cos A\\ \Leftrightarrow {10^2} = {b^2} + {c^2} – 2bc.\cos 60\\ \Leftrightarrow 100 = {b^2} + {c^2} – bc\,\,\left( 2 \right)\end{array}\) Từ (1) và (2) suy ra \(\begin{array}{l}\left\{ \begin{array}{l}3bc = 10\left( {10 + b + c} \right)\\100 = {b^2} + {c^2} – bc\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}3bc = 10\left( {10 + b + c} \right)\\100 = {\left( {b + c} \right)^2} – 3bc\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}3bc = 100 + 10\left( {b + c} \right)\\3bc = {\left( {b + c} \right)^2} – 100\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}100 + 10\left( {b + c} \right) = {\left( {b + c} \right)^2} – 100\\3bc = 100 + 10\left( {b + c} \right)\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}{\left( {b + c} \right)^2} – 10\left( {b + c} \right) – 200 = 0\\3bc = 100 + 10\left( {b + c} \right)\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}b + c = 20\\b + c = – 10\left( {loai} \right)\end{array} \right.\\3bc = 100 + 10\left( {b + c} \right)\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}b + c = 20\\bc = 100\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}b = 10\\c = 10\end{array} \right.\end{array}\) Vậy \(R = \dfrac{{10\sqrt 3 }}{3},b = 10,c = 10\) Bình luận
Đáp án:
\(R = \dfrac{{10\sqrt 3 }}{3},b = 10,c = 10\)
Giải thích các bước giải:
Áp dụng định lí sin:
\(\dfrac{a}{{\sin A}} = 2R \Leftrightarrow R = \dfrac{a}{{2\sin A}} = \dfrac{{10}}{{2\sin {{60}^0}}} = \dfrac{{10\sqrt 3 }}{3}\)
Ta có:
\(\begin{array}{l}S = \dfrac{{abc}}{{4R}} = pr \Leftrightarrow \dfrac{{10bc}}{{4.\dfrac{{10\sqrt 3 }}{3}}} = \dfrac{{10 + b + c}}{2}.\dfrac{{5\sqrt 3 }}{3}\\ \Leftrightarrow 60bc = \dfrac{{40\sqrt 3 }}{3}.5\sqrt 3 \left( {10 + b + c} \right)\\ \Leftrightarrow 60bc = 200\left( {10 + b + c} \right)\\ \Leftrightarrow 3bc = 10\left( {10 + b + c} \right)\,\,\left( 1 \right)\end{array}\)
Áp dụng định lý cos:
\(\begin{array}{l}{a^2} = {b^2} + {c^2} – 2bc\cos A\\ \Leftrightarrow {10^2} = {b^2} + {c^2} – 2bc.\cos 60\\ \Leftrightarrow 100 = {b^2} + {c^2} – bc\,\,\left( 2 \right)\end{array}\)
Từ (1) và (2) suy ra
\(\begin{array}{l}\left\{ \begin{array}{l}3bc = 10\left( {10 + b + c} \right)\\100 = {b^2} + {c^2} – bc\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}3bc = 10\left( {10 + b + c} \right)\\100 = {\left( {b + c} \right)^2} – 3bc\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}3bc = 100 + 10\left( {b + c} \right)\\3bc = {\left( {b + c} \right)^2} – 100\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}100 + 10\left( {b + c} \right) = {\left( {b + c} \right)^2} – 100\\3bc = 100 + 10\left( {b + c} \right)\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}{\left( {b + c} \right)^2} – 10\left( {b + c} \right) – 200 = 0\\3bc = 100 + 10\left( {b + c} \right)\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}b + c = 20\\b + c = – 10\left( {loai} \right)\end{array} \right.\\3bc = 100 + 10\left( {b + c} \right)\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}b + c = 20\\bc = 100\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}b = 10\\c = 10\end{array} \right.\end{array}\)
Vậy \(R = \dfrac{{10\sqrt 3 }}{3},b = 10,c = 10\)