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8. Introduction to Trigonometry
hard
Given $15 \cot A =8,$ find $\sin A$ and $\sec A .$
Option A
Option B
Option C
Option D
Solution
Consider a right-angled triangle, right-angled at $B.$
$\cot A=\frac{\text { Side adjacent to } \angle A }{\text { Side opposite to } \angle A }$
$=\frac{A B}{B C}$
It is given that,
$\cot A=\frac{8}{15}$
$\frac{A B}{B C}=\frac{8}{15}$
Let $AB$ be $8 k$. Therefore, $BC$ will be $15 k ,$ where $k$ is a positive integer.
Applying Pythagoras theorem in $\triangle ABC ,$ we obtain
$AC ^{2}= AB ^{2}+ BC ^{2}$
$=(8 k)^{2}+(15 k)^{2}$
$=64 k^{2}+225 k^{2}$
$=289 k^{2}$
$AC =17 k$
$\sin A=\frac{\text { Side opposite to } \angle A }{\text { Hypotenuse }}=\frac{ BC }{ AC }$
$=\frac{15 k}{17 k}=\frac{15}{17}$
$\sec A=\frac{\text { Hypotenuse }}{\text { Side adjacent to } \angle A }$
$=\frac{ AC }{ AB }=\frac{17}{8}$
Standard 10
Mathematics
