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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