Question 25

ΔDEF is right angled at E. If $$cosec D = \frac{5}{4}$$, then what is the value of $$cosec F$$ ?

Solution

Given : $$\cosec D$$ = $$\frac{5}{4}$$

Also, $$\cosec D=\frac{DF}{EF}=\frac{5}{4}$$

Let DF = 5 cm and EF = 4 cm

Thus, in $$\triangle$$ DEF, => $$(DE)^2=(DF)^2-(EF)^2$$

=> $$(DE)^2=(5)^2-(4)^2$$

=> $$(DE)^2=25-16=9$$

=> $$DE=\sqrt{9}=3$$ cm

To find : $$\cosec F=\frac{DF}{DE}$$

= $$\frac{5}{3}$$

=> Ans - (A)

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