Ta có:

\(AM^2=BM\cdot CM\)

\(\Rightarrow CM=\dfrac{AM^2}{BM}=\dfrac{6^2}{2}=18\left(m\right)\)

Quãng đường từ bờ sông đền khóm hoa là:

\(BC=CM+BM=18+2=20\left(m\right)\)

Ta có:

\(DE^2=AE\cdot CE\)

\(\Rightarrow CE=\dfrac{DE^2}{AE}=\dfrac{2,25^2}{1,5}=3,375\left(m\right)\)

Chiều cao của cây là:
\(AC=AE+CE=1,5+3,375=4,875\left(m\right)\)

a, Do ABCD là hình chữ nhật 

⇒ AC = BD =\(\sqrt{AB^2+BC^2}=\sqrt{8^2+15^2}=17\)

B, Áp dụng hệ thức lượng vào tam giác vuông ABD có: 

\(AH=\dfrac{AD\cdot AB}{BD}=\dfrac{8\cdot15}{17}=\dfrac{120}{17}\)

Bài 38:

Ta có: \(AH^2=BH\cdot CH\)

\(\Rightarrow CH=\dfrac{AH^2}{BH}=\dfrac{12^2}{9}=16\left(cm\right)\)

Mà: \(BC=BH+CH=9+16=25\left(cm\right)\)

\(\Rightarrow S_{ABC}=\dfrac{1}{2}\cdot AH\cdot BC=\dfrac{1}{2}\cdot12\cdot25=150\left(cm^2\right)\)

Bài 39: 

Ta có: \(\dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}\)

\(\Rightarrow AH^2=\dfrac{AB^2\cdot AC^2}{AB^2+AC^2}\Leftrightarrow12^2=\dfrac{AB^2\cdot20^2}{AB^2+20^2}\)

\(\Leftrightarrow AB=15\left(cm\right)\)

Áp dụng định lý Py-ta-go ta có:

\(BC=\sqrt{AC^2+AB^2}=\sqrt{20^2+15^2}=25\left(cm\right)\)

\(\Rightarrow S_{ABC}=\dfrac{1}{2}\cdot AH\cdot BC=\dfrac{1}{2}\cdot12\cdot25=150\left(cm^2\right)\)

1:

a: BD=căn 8^2+15^2=17(cm)

b: Xét ΔABD vuông tại A có AH là đường cao

nên AH*BD=AB*AD

=>AH*17=8*15=120

=>AH=120/17(cm)

c: Xét ΔHDK vuông tại H và ΔHIB vuông tại H có

góc HDK=góc HIB

=>ΔHDK đồng dạng với ΔHIB

=>HD/HI=HK/HB

=>HI*HK=HD*HB=AH^2

\(A=\left(\dfrac{1}{\sqrt{x-1}}+\dfrac{1}{\sqrt{x+1}}\right)^2\cdot\dfrac{x^2-1}{2}-\sqrt{x^2-1}\\ \Rightarrow A=\left(\dfrac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x^2-1}}\right)^2\cdot\dfrac{x^2-1}{2}-\sqrt{x^2-1}\\ \Rightarrow A=\dfrac{\left(\sqrt{x+1}+\sqrt{x-1}\right)^2}{2}-\sqrt{x^2-1}\\ \Rightarrow A=\dfrac{2x+2\sqrt{x^2-1}-2\sqrt{x^2-1}}{2}\\ \Rightarrow A=x\)

a) Ta có:

\(AC^2=13^2=169\)

\(AB^2+BC^2=5^2+12^2=25+144=169\)

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

\(\Rightarrow\Delta ABC\) vuông tại B (theo định lý Pytago đảo)

b) Ta có:

\(sinA=cosC=\dfrac{BC}{AC}=\dfrac{12}{13}\)

\(cosA=sinC=\dfrac{AB}{AC}=\dfrac{5}{13}\)

\(tanA=cotC=\dfrac{BC}{AB}=\dfrac{12}{5}\)

\(cotA=tanC=\dfrac{AB}{BC}=\dfrac{5}{12}\)

a. \(\Delta ABC\) có

\(AB^2+BC^2=5^2+12^2=169\)

\(AC^2=13^2=169\)

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

\(\Rightarrow\Delta ABC\perp tại.B\)

b. \(sin.A=\dfrac{BC}{AC}=\dfrac{12}{13}\\ cos.A=\dfrac{AB}{AC}=\dfrac{5}{13}\\ tan.A=\dfrac{BC}{AB}=\dfrac{12}{5}\\ cot.A=\dfrac{AB}{BC}=\dfrac{5}{12}\)

\(sin.C=\dfrac{AB}{AC}=\dfrac{5}{13}\\ cos.C=\dfrac{BC}{AC}=\dfrac{12}{13}\\ tan.C=\dfrac{AB}{BC}=\dfrac{5}{12}\\ cot.C=\dfrac{BC}{AB}=\dfrac{12}{5}\)

\(tanx=\dfrac{4}{3}\)

\(\Rightarrow cotx=\dfrac{1}{tanx}=\dfrac{1}{\dfrac{4}{3}}=\dfrac{3}{4}\)

\(1+tan^2x=\dfrac{1}{cos^2x}\)

\(\Rightarrow cos^2x=\dfrac{1}{1+tan^2x}\)

\(=\dfrac{1}{1+\left(\dfrac{4}{3}\right)^2}=\dfrac{1}{1+\dfrac{16}{9}}=\dfrac{1}{\dfrac{25}{9}}=\dfrac{9}{25}\)

\(\Rightarrow cosx=\dfrac{3}{5}\)

\(sin^2x+cos^2x=1\)

\(\Rightarrow sin^2x=1-cos^2x=1-\left(\dfrac{3}{5}\right)^2=1-\dfrac{9}{25}=\dfrac{16}{25}\)

\(\Rightarrow sinx=\dfrac{4}{5}\)

Có \(tan.\alpha=\dfrac{4}{3}\)

Mà \(tan.\alpha.cot.\alpha=1\)

\(\Rightarrow cot.\alpha=1:\dfrac{4}{3}=\dfrac{3}{4}\)

Lại có \(sin^2\alpha+cos^2\alpha=1\\ \Leftrightarrow sin^2\alpha=1-cos^2\alpha\\ \Leftrightarrow sin\alpha=\sqrt{1-cos^2\alpha}\)

Vì \(tan.\alpha=\dfrac{sin.\alpha}{cos.\alpha}\)

\(\Leftrightarrow\dfrac{4}{3}=\dfrac{\sqrt{1-cos^2\alpha}}{cos.\alpha}\)

\(\Leftrightarrow\dfrac{4}{3}=\dfrac{1-cos^2\alpha}{cos^2\alpha}\\ \Leftrightarrow4.cos^2\alpha=3.\left(1-cos^2\alpha\right)\\ \Leftrightarrow4.cos^2\alpha=3-3cos^2\alpha\\ \Leftrightarrow cos.\alpha=\dfrac{\sqrt{21}}{7}\)

\(\Rightarrow sin.\alpha=\sqrt{1-\left(\dfrac{\sqrt{21}}{7}\right)^2}=\dfrac{4}{7}\)