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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