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Question 25
∆ABC is right angled at B. If cos A = $$\frac{8}{17}$$, then what is the value of cot C ?
Solution
Given : $$\cos A$$ = $$\frac{8}{17}$$
Also, $$\cos A=\frac{AB}{AC}=\frac{8}{17}$$
Let AB = 8 cm and AC = 17 cm
Thus, in $$\triangle$$ ABC,
=> $$(BC)^2=(AC)^2-(AB)^2$$
=> $$(BC)^2=(17)^2-(18)^2$$
=> $$(BC)^2=289-64=225$$
=> $$BC=\sqrt{225}=15$$ cm
To find : $$\cot C=\frac{BC}{AB}$$
= $$\frac{15}{8}$$
=> Ans - (A)
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