Question 25
ΔUVW is right angled at V. If $$sec U = \frac{5}{3}$$, then what is the value of $$tan W$$ ?
Solution
Given : $$\sec U$$ = $$\frac{5}{3}$$
Also, $$\sec U=\frac{UW}{UV}=\frac{5}{3}$$
Let UW = 5 cm and UV = 3 cm
Thus, in $$\triangle$$ UVW, => $$(VW)^2=(UW)^2-(UV)^2$$
=> $$(VW)^2=(5)^2-(3)^2$$
=> $$(VW)^2=25-9=16$$
=> $$VW=\sqrt{16}=4$$ cm
To find : $$\tan W=\frac{UV}{VW}$$
= $$\frac{3}{4}$$
=> Ans - (A)
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