question_answer

If \[x\,\cos \,A=8\]and \[15\,\cos ec\,A=8\,\sec \,A\], then the value of x is

A) 20

B) 16

C) 17

D) 13

Correct Answer: C

Solution :

Given, \[x\,\cos \,A=8\]

\[\cos \,A=\frac{8}{x}\]

And      \[15\,\cos ec\,A=8\,\sec \,A\]

\[\cos \,A=\frac{8}{x}\]

\[=\frac{Base}{Hypotenuse}=\frac{AB}{AC}\]

Let  \[AB=8k,\,AC=xk\]

Apply Pythagoras theorem,

\[A{{B}^{2}}+B{{C}^{2}}=A{{C}^{2}}\]

\[\Rightarrow {{\left( 8k \right)}^{2}}+B{{C}^{2}}={{\left( xk \right)}^{2}}\]

\[\Rightarrow BC=\left( \sqrt{{{x}^{2}}-{{8}^{2}}} \right)k\]

\[\Rightarrow 15\cos ecA=8\sec A\]

\[\Rightarrow \frac{15\times xk}{k\sqrt{{{x}^{2}}-{{8}^{2}}}}=\frac{8xk}{8k}\]

\[\Rightarrow \sqrt{{{x}^{2}}-{{8}^{2}}}=15\]

On squaring both side, we get

\[{{x}^{2}}-{{8}^{2}}={{\left( 15 \right)}^{2}}\]

\[\Rightarrow {{x}^{2}}=225+64\]

\[\Rightarrow {{x}^{2}}=289\]

\[\Rightarrow x=17\]