Question
Given 15 cot A = 8 find sin A and sec A.
✓
Answer
Let us draw a right triangle ABC.
15 cot A = 8 ...... Given
$\Rightarrow \cot A = \frac { 8 } { 15 } \Rightarrow \frac { A B } { B C } = \frac { 8 } { 15 }$
$\Rightarrow \frac { A B } { 8 } = \frac { B C } { 15 } = k ( 5 a y )$
where k is a positive number
$\Rightarrow A B = 8 k$
BC = 15k
By using the Pythagoras theorem, we have
$A C ^ { 2 } = A B ^ { 2 } + B C ^ { 2 }$
$\Rightarrow A C ^ { 2 } = ( 8 k ) ^ { 2 } + ( 15 k ) ^ { 2 } \Rightarrow A C ^ { 2 } = 64 k ^ { 2 } + 225 k ^ { 2 }$
$\Rightarrow A C = \sqrt { 289 k ^ { 2 } } \Rightarrow A C = 17 k$
Now, $\sin A = \frac { B C } { A C } = \frac { 15 k } { 17 k } = \frac { 15 } { 17 }$
and, $\sec A = \frac { A C } { A B } = \frac { 17 k } { 8 k } = \frac { 17 } { 8 }$
15 cot A = 8 ...... Given
$\Rightarrow \cot A = \frac { 8 } { 15 } \Rightarrow \frac { A B } { B C } = \frac { 8 } { 15 }$
$\Rightarrow \frac { A B } { 8 } = \frac { B C } { 15 } = k ( 5 a y )$
where k is a positive number
$\Rightarrow A B = 8 k$
BC = 15k
By using the Pythagoras theorem, we have
$A C ^ { 2 } = A B ^ { 2 } + B C ^ { 2 }$
$\Rightarrow A C ^ { 2 } = ( 8 k ) ^ { 2 } + ( 15 k ) ^ { 2 } \Rightarrow A C ^ { 2 } = 64 k ^ { 2 } + 225 k ^ { 2 }$
$\Rightarrow A C = \sqrt { 289 k ^ { 2 } } \Rightarrow A C = 17 k$
Now, $\sin A = \frac { B C } { A C } = \frac { 15 k } { 17 k } = \frac { 15 } { 17 }$
and, $\sec A = \frac { A C } { A B } = \frac { 17 k } { 8 k } = \frac { 17 } { 8 }$
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