\(\frac{3}{4}=tanB=\frac{AC}{AB}\Rightarrow AC=\frac{3}{4}AB\)

\(BC^2=AB^2+AC^2\)(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

Trả lời giúp mk vs 

Ta có: AB; AC tỉ lệ với 8; 15

=> AB = 815815 AC

Ta có: tam giác ABC vuông tại A

=> BC2 = AB2 + AC2

=> 1022 =( 815815AC)2 + AC2

=> 10404 = 6422564225 . AC2 + AC2

=> 10404 = AC2. (64225+164225+1)

=> 10404 = AC2 . 289225289225

=> AC2 = 10404 : 289225289225 = 8100

=> AC2 = 902

=> AC = 90 cm

Ta có: AB = 815815AC

=> AB = 815.90815.90=48 cm

Vậy AB = 48 cm

AC = 90 cm

T I C K mk nha

Ta có: \(\dfrac{AB}{AC}=\dfrac{8}{15}\)(gt)

nên \(AB=\dfrac{8}{15}\cdot AC\)

Áp dụng định lí Pytago vào ΔABC vuông tại A, ta được:

\(BC^2=AB^2+AC^2\)

\(\Leftrightarrow\left(\dfrac{8}{15}\cdot AC\right)^2+AC^2=102^2\)

\(\Leftrightarrow\dfrac{64}{225}AC^2+AC^2=102^2\)

\(\Leftrightarrow\dfrac{289}{225}AC^2=102^2\)

\(\Leftrightarrow AC^2=102^2:\dfrac{289}{225}=8100\)

hay AC=90(cm)

Ta có: \(AB=AC\cdot\dfrac{8}{15}\)(cmt)

nên \(AB=90\cdot\dfrac{8}{15}=48\left(cm\right)\)

Vậy: AC=90cm; AB=48cm

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