Bài 4: Một số hệ thức về cạnh và góc trong tam giác vuông

Câu 4:

\(a,\sin B=\dfrac{AC}{BC}=\dfrac{12}{13};\cos B=\dfrac{AB}{BC}=\dfrac{5}{13};\tan B=\dfrac{AC}{AB}=\dfrac{12}{5};\cot B=\dfrac{AB}{AC}=\dfrac{5}{12}\\ b,\text{Áp dụng HTL: }\left\{{}\begin{matrix}AH=\dfrac{AB\cdot AC}{BC}=\dfrac{60}{13}\left(cm\right)\\BH=\dfrac{AB^2}{BC}=\dfrac{25}{13}\left(cm\right)\end{matrix}\right.\\ \sin B=\dfrac{12}{13}\approx67^0\\ \Rightarrow\widehat{B}\approx67^0\\ \Rightarrow\widehat{HAB}=90^0-\widehat{B}\approx23^0\)

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Bài đâu?

\(a,VT=\left(\sin^252^0+\sin^238^0\right)-\left(\tan37^0-\cot53^0\right)+\dfrac{\tan42^0}{\tan42^0}\\ =\left(\sin^252^0+\cos^252^0\right)-\left(\tan37^0-\tan37^0\right)+1\\ =1-0+1=2=VP\\ c,VT=\dfrac{2\cos^2\alpha-\sin^2\alpha-\cos^2\alpha}{\sin\alpha+\cos\alpha}=\dfrac{\cos^2\alpha-\sin^2\alpha}{\sin\alpha+\cos\alpha}\\ =\dfrac{\left(\cos\alpha-\sin\alpha\right)\left(\cos\alpha+\sin\alpha\right)}{\cos\alpha+\sin\alpha}=\cos\alpha-\sin\alpha=VP\\ b,VT=\cos^2\alpha+\cos^2\alpha\cdot\dfrac{\sin^2\alpha}{\cos^2\alpha}=\cos^2\alpha+\sin^2\alpha=1=VP\)

\(a,AB=BK+KA=25\)

Áp dụng HTL: \(\left\{{}\begin{matrix}CK=\sqrt{AK\cdot BK}=12\\CA=\sqrt{AK\cdot BA}=20\\CB=\sqrt{BK\cdot BA}=15\end{matrix}\right.\)

\(b,\) Sửa: Tính \(\widehat{BCK}\)

Ta có \(\sin\widehat{BCK}=\dfrac{BK}{BC}=\dfrac{3}{5}\approx\sin37^0\)

Do đó \(\widehat{BCK}\approx37^0\)

\(c,\) Vì CH là p/g nên \(\widehat{BCH}=\dfrac{1}{2}\widehat{BCA}=45^0\)

\(\Rightarrow\widehat{KCH}=\widehat{BCH}-\widehat{BCK}\approx45^0-37^0=8^0\)

Xét tg KCH vuông tại K thì \(\widehat{CHK}=90^0-\widehat{KCH}=82^0\)

a: AC=8cm

AH=4,8cm

BH=3,6cm

CH=6,4cm

Áp dụng hệ thức lượng:

\(AB^2=BH.BC\Rightarrow BC=\dfrac{AB^2}{BH}=10\left(cm\right)\)

\(AC=\sqrt{BC^2-AB^2}=8\left(cm\right)\)

\(AH.BC=AB.AC\Rightarrow AH=\dfrac{AB.AC}{BC}=4,8\left(cm\right)\)