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\(\frac{3}{4}=tanB=\frac{AC}{AB}\Rightarrow AC=\frac{3}{4}AB\)
\(BC^2=AB^2+AC^2\)(định lí Pythagore)
\(=AB^2+\frac{9}{16}AB^2=\frac{25}{16}AB^2\)
\(\Rightarrow AB^2=10^2.\frac{16}{25}\Rightarrow AB=8\left(cm\right)\)
\(\Rightarrow AC=\frac{3}{4}AB=6\left(cm\right)\)
\(AC=\sqrt{BC^2-AB^2}=10\sqrt{3}\left(cm\right)\left(pytago\right)\\ \sin B=\dfrac{AC}{BC}=\dfrac{\sqrt{3}}{2}=\sin60^0\Rightarrow\widehat{B}=60^0\\ \widehat{C}=90^0-\widehat{B}=30^0\\ 2,\sin B\cdot\tan B=\dfrac{AC}{AB}\cdot\dfrac{AC}{BC}=\dfrac{AC^2}{AB\cdot BC}=\dfrac{HC\cdot BC}{AB\cdot BC}=\dfrac{HC}{AB}\\ 3,\dfrac{CI}{IB}=\dfrac{AC}{AB}=\sqrt{3}\Leftrightarrow CI=\sqrt{3}IB\\ CI+IB=BC=20\\ \Rightarrow\left(\sqrt{3}+1\right)IB=20\Leftrightarrow IB=\dfrac{20}{\sqrt{3}+1}=10\sqrt{3}-10\left(cm\right)\\ HB=\dfrac{AB^2}{BC}=5\left(cm\right)\left(HTL\right)\\ IH=IB-HB=10\sqrt{3}-15\left(cm\right)\)
ta có:cosB=\(\dfrac{AB}{BC}=\dfrac{5}{13}\)⇒BC=39
AC=\(\sqrt{BC^2-AB^2}=\sqrt{39^2-15^2}\)=36
Áp dụng định lí Pi-ta-go vào tam giác vuông ABC, ta có:
B C 2 = A B 2 + A C 2 ⇒ A B 2 = B C 2 - A C 2
\(1,HC=\dfrac{AH^2}{BH}=\dfrac{256}{9}\\ \Rightarrow AB=\sqrt{BH\cdot BC}=\sqrt{\left(\dfrac{256}{9}+9\right)9}=\sqrt{337}\\ 2,BC=\sqrt{AB^2+AC^2}=10\left(cm\right)\\ \Rightarrow BH=\dfrac{AB^2}{BC}=6,4\left(cm\right)\\ 3,AC=\sqrt{BC^2-AB^2}=9\\ \Rightarrow CH=\dfrac{AC^2}{BC}=5,4\\ 4,AC=\sqrt{BC\cdot CH}=\sqrt{9\left(6+9\right)}=3\sqrt{15}\\ 5,AC=\sqrt{BC^2-AB^2}=4\sqrt{7}\left(cm\right)\\ \Rightarrow AH=\dfrac{AB\cdot AC}{BC}=3\sqrt{7}\left(cm\right)\\ 6,AC=\sqrt{BC\cdot CH}=\sqrt{12\left(12+8\right)}=4\sqrt{15}\left(cm\right)\)
Áp dụng định lí pi ta go
=> AB2 + AC2 = 289
Mà \(\dfrac{AB}{AC}\) = \(\dfrac{8}{15}\)=> (\(\dfrac{AB}{AC}\))2 = \(\dfrac{64}{225}\)
=> AC2=225 => AC = 15 => AB = 8
Ta có: AB.AC=BC . AH
=> AH = 120/17=7.06
=>BH = 3.76
=> CH = 13.24
Đúng thì like giúp mik nha bạn. Thx bạn
Bài 2:
Ta có: \(\dfrac{HB}{HC}=\dfrac{1}{3}\)
nên HC=3HB
Ta có: \(AH^2=HB\cdot HC\)
\(\Leftrightarrow HB^2=48\)
\(\Leftrightarrow HB=4\sqrt{3}\left(cm\right)\)
\(\Leftrightarrow BC=4\cdot HB=16\sqrt{3}\left(cm\right)\)
Bài 1:
ta có: \(AB=\dfrac{1}{2}AC\)
\(\Leftrightarrow\dfrac{HB}{HC}=\dfrac{1}{4}\)
\(\Leftrightarrow HC=4HB\)
Ta có: \(AH^2=HB\cdot HC\)
\(\Leftrightarrow HB=1\left(cm\right)\)
\(\Leftrightarrow HC=4\left(cm\right)\)
hay BC=5(cm)
Xét ΔBAC vuông tại A có AH là đường cao ứng với cạnh huyền BC
nên \(\left\{{}\begin{matrix}AB^2=HB\cdot BC\\AC^2=HC\cdot BC\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}AB=\sqrt{5}\left(cm\right)\\AC=2\sqrt{5}\left(cm\right)\end{matrix}\right.\)
AC=AB.tg B
AC= 30.\(\dfrac{8}{15}\)
AC= 16cm
BC2=AB2+AC2
BC2 = 900+256=1156
BC=34cm
Ta có: \(tgB=\dfrac{8}{15}\Rightarrow\dfrac{AC}{AB}=\dfrac{8}{15}\Rightarrow AC=\dfrac{8AB}{15}=\dfrac{8.30}{15}=16\left(cm\right)\)
Xét tam giác ABC vuông tại A có:
\(BC^2=AB^2+AC^2\) ( định lý Pytago)
\(\Rightarrow BC=\sqrt{AB^2+AC^2}=\sqrt{30^2+16^2}=34\left(cm\right)\)