In a right-angled triangle XYZ,right-angled a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given: $XZ - YZ = 2$ and $XY = 2\sqrt{6}$.In right-angled triangle $XYZ$,by Pythagoras theorem:$XY^2 + YZ^2 = XZ^2$$(2\sqrt{6})^2 = XZ^2 - YZ^2$$2

Vedclass · Accepted Answer

$\sqrt{6}$

Vedclass · Answer

(B) Given: $XZ - YZ = 2$ and $XY = 2\sqrt{6}$.In right-angled triangle $XYZ$,by Pythagoras theorem:$XY^2 + YZ^2 = XZ^2$$(2\sqrt{6})^2 = XZ^2 - YZ^2$$24 = (XZ - YZ)(XZ + YZ)$Since $XZ - YZ = 2$,we have:$24 = 2(XZ + YZ)$$XZ + YZ = 12$Now,we have a system of two linear equations:$1$) $XZ - YZ = 2$$2$) $XZ + YZ = 12$Adding equations $(1)$ and $(2)$:$2XZ = 14 \Rightarrow XZ = 7$Substituting $XZ = 7$ in equation $(1)$:$7 - YZ = 2 \Rightarrow YZ = 5$Now,for angle $X$:$\sec X = \frac{	ext{Hypotenuse}}{	ext{Adjacent side}} = \frac{XZ}{XY} = \frac{7}{2\sqrt{6}}$$	an X = \frac{	ext{Opposite side}}{	ext{Adjacent side}} = \frac{YZ}{XY} = \frac{5}{2\sqrt{6}}$Therefore,$\sec X + 	an X = \frac{7}{2\sqrt{6}} + \frac{5}{2\sqrt{6}} = \frac{12}{2\sqrt{6}} = \frac{6}{\sqrt{6}} = \sqrt{6}$.

In a right-angled triangle $XYZ$,right-angled at $Y$,if $XY = 2\sqrt{6}$ and $XZ - YZ = 2$,then $\sec X + \tan X$ is

Similar Questions

If $y = 3\,sin\,x + 4\,cos\,x$,then find the maximum value of $y$.

$\sin 12^\circ \sin 48^\circ \sin 54^\circ = $

If $\sin \theta = -\frac{4}{5}$ and $\theta$ lies in the third quadrant,then $\cos \frac{\theta}{2} = $

$2\sin^2 \beta + 4\cos(\alpha + \beta)\sin \alpha \sin \beta + \cos 2(\alpha + \beta) = $

The radius of the circle whose arc of length $15 \ cm$ makes an angle of $3/4$ radian at the centre is ..... $cm$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module