Bài 4. Cho tam giác ABC vuông tại A, đường cao AH, biết AB = 6cm, HC = 4cm. Tính sinB, cosB, tanC 24/08/2021 Bởi Arianna Bài 4. Cho tam giác ABC vuông tại A, đường cao AH, biết AB = 6cm, HC = 4cm. Tính sinB, cosB, tanC
Đặt $BH=x$ khi đó ta có: $AB^2=BH.BC=x.(BH+CH)=x.(x+4)$ $⇒x.(x+4)=6^2$$⇒x^2+4x-36=0$ $⇒(x+2)^2=40$ $⇔$\(\left[ \begin{array}{l}x+2=\sqrt[]{40}\\x+2=-\sqrt[]{40}\end{array} \right.\) Mà $x+2>0$ $⇒x=2.\sqrt[]{10}-2$ $⇒BH=2.\sqrt[]{10}-2$ $⇒BC=2.\sqrt[]{10}-2+4=2\sqrt[]{10}+2$ $⇒AC=\sqrt[]{BC^2-AB^2}=\sqrt[]{(2\sqrt[]{10}+2)^2-6^2}=5,77(cm)$ $⇒sinB=\dfrac{AC}{BC}=\dfrac{5,77}{2.\sqrt[]10+2}$ $cosB=\dfrac{AB}{BC}=\dfrac{6}{2.\sqrt[]10+2}$ $tanB=\dfrac{\dfrac{5,77}{2.\sqrt[]10+2}}{\dfrac{6}{2.\sqrt[]10+2}}=\dfrac{5,77}{6}$ $⇒cotB=\dfrac{6}{5,77}$ $⇒tanC=\dfrac{6}{5,77}$ Bình luận
Đặt $BH=x$ khi đó ta có:
$AB^2=BH.BC=x.(BH+CH)=x.(x+4)$
$⇒x.(x+4)=6^2$
$⇒x^2+4x-36=0$
$⇒(x+2)^2=40$
$⇔$\(\left[ \begin{array}{l}x+2=\sqrt[]{40}\\x+2=-\sqrt[]{40}\end{array} \right.\)
Mà $x+2>0$
$⇒x=2.\sqrt[]{10}-2$
$⇒BH=2.\sqrt[]{10}-2$
$⇒BC=2.\sqrt[]{10}-2+4=2\sqrt[]{10}+2$
$⇒AC=\sqrt[]{BC^2-AB^2}=\sqrt[]{(2\sqrt[]{10}+2)^2-6^2}=5,77(cm)$
$⇒sinB=\dfrac{AC}{BC}=\dfrac{5,77}{2.\sqrt[]10+2}$
$cosB=\dfrac{AB}{BC}=\dfrac{6}{2.\sqrt[]10+2}$
$tanB=\dfrac{\dfrac{5,77}{2.\sqrt[]10+2}}{\dfrac{6}{2.\sqrt[]10+2}}=\dfrac{5,77}{6}$
$⇒cotB=\dfrac{6}{5,77}$
$⇒tanC=\dfrac{6}{5,77}$