Bài 1:
Xét ΔABC \(\left(\widehat{BAC}=90^o\right)\)theo định lí py - ta - go ta có:
\(AC=\sqrt{BC^2-AB^2}=\sqrt{10^2-8^2}=6\)
Xét ΔABC \(\left(\widehat{BAC}=90^o\right)\)theo hệ thức lượng ta có:
\(\frac{1}{AH^2}=\frac{1}{AB^2}+\frac{1}{AC^2}=\frac{1}{8^2}+\frac{1}{6^2}=\frac{25}{576}\)
=>\(AH^2=\frac{576}{25}=23,04=>AH=\sqrt{23,04}=4,8\)
Xét ΔAHC \(\left(\widehat{AHC}=90^o\right)\)theo định lí py - ta - go ta có:
\(HC=\sqrt{AC^2-AH^2}=\sqrt{6^2-4,8^2}=3,6\)
=> HB = BC - HC = 10 - 3,6 = 6,4
Ta có: Sin B = \(\frac{AH}{AB}=\frac{4,8}{8}=>\widehat{B}\approx37^o\)
=> Sin C = \(\frac{AH}{AC}=\frac{4,8}{6}=>\widehat{C}\approx53^o\)
\(\widehat{BAH}=90^o-37^o=53^o\)
\(\widehat{HAC}=90^o-53^o=37^o\)
Bài 2:
Xét ΔABC \(\left(\widehat{ABC}=90^o\right)\)theo định lí py - ta - go ta có:
\(BC=\sqrt{AC^2-AB^2}=\sqrt{15^2-9^2}=12cm\)
Xét ΔABC \(\left(\widehat{ABC}=90^o\right)\)theo hệ thức lượng ta có:
\(\frac{1}{BM^2}=\frac{1}{AB^2}+\frac{1}{BC^2}=\frac{1}{9^2}+\frac{1}{12^2}=\frac{25}{1296}=>BM^2=\frac{1296}{25}=51,84\)
\(=>BM=\sqrt{51,84}=7,2cm\)
Xét ΔAMB \(\left(\widehat{AMB}=90^o\right)\)theo định lí py - ta - go ta có:
\(AM=\sqrt{AB^2-BM^2}=\sqrt{9^2-7,2^2}=5,4cm\)
=> MC = AC - AM = 15 - 5,4 = 9,6cm
Ta có: Sin A = \(\frac{BM}{AB}=\frac{7,2}{9}=>\widehat{B}\approx53^o\)
Sin C = \(\frac{BM}{BC}=\frac{7,2}{12}=>\widehat{C}\approx37^o\)
\(\widehat{ABM}=90^o-53^o=37^o\)
\(\widehat{CBM}=90^o-37^o=53^o\)
