Question 139

$$\angle Y$$ is the right angle of the triangle $$XYZ$$. If $$XY=2\sqrt{6}$$ cm and $$XZ-YZ=2$$ cm, then the value of $$(secX + tanX)$$ is:

Solution

Given : $$XY=2\sqrt{6}$$ cm and $$XZ-YZ=2$$

To find : $$(secX + tanX)$$ = ?

Solution : In $$\triangle$$ XYZ,

=> $$(XY)^2=(XZ)^2-(YZ)^2$$

=> $$(2\sqrt6)^2=(XZ-YZ)(XZ+YZ)$$

=> $$(2)(XZ+YZ)=24$$

=> $$(XZ+YZ)=\frac{24}{2}=12$$ -------------(i)

$$\therefore$$ $$(secX + tanX)$$

= $$(\frac{XZ}{XY})+(\frac{YZ}{XY})=\frac{(XZ+YZ)}{XY}$$

= $$\frac{12}{2\sqrt6}=\sqrt6$$

=> Ans - (D)

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