Question
If tan A $= 15$, find sin A - cos A.
✓
Answer
Given: $8\tan\text{A}=15$ Therefore, $\tan\text{A}=\frac{15}{8}\ \dots(1)$
To find: $\sin\text{A}-\cos\text{A}$
Now we know $\tan\theta$ is defined as follows
$\tan\text{A}=\frac{\text{Perpendicular side oppsite to }\angle\text{A}}{\text{Base side adjacement to }\angle\text{A}}\ \dots(2)$
Now by comparing equation $(1)$ and $(2)$ We get, Perpendicular side opposite to $\angle\text{A}=15$ Base side adjacement to $\angle\text{A}=8$ Therefore triangle representing angle A is as shown below
Side AC is unknown and can be found using Pythagoras theorem Therefore, $AC^2 = AB^2 + BC^2$
Now by substituting the value of known sides from figure (a) We get,
$AC^2 = 15^2 + 8^2 = 225 + 64 = 289$
Now by taking square root on both sides We get, $\text{AC}=\sqrt{289}$ $=17$ Therefore Hypotenuse side AC = 17 …… (3)
Now we know, sin A is defined as follows
$\sin\text{A}=\frac{\text{Peependicular side opposite to}\angle\text{A}}{\text{Hypotenuse}}$ Therefore from figure (a) and equation (3) We get, $\sin\text{A}=\frac{\text{BC}}{\text{AC}}$ $=\frac{15}{17}$ $\sin\text{A}=\frac{15}{17}\ \dots(4)$
Now we know, cos A is defined as follows
$\cos\text{A}=\frac{\text{Base side adjacement to}\angle\text{A}}{\text{Hypotenuse}}$
Therefore from figure (a) and equation (3) We get, $\cos\text{A}=\frac{\text{AB}}{\text{AC}}$
$=\frac{8}{17}$ $\cos\text{A}=\frac{8}{17}\ \dots(5)$
Now we need to find the value of expression $\sin\text{A}-\cos\text{A}$
Therefore by substituting the value of sin A and cos A from equation $(4)$ and $(5)$ respectively,
we get, $\sin\text{A}-\cos\text{A}=\frac{15}{17}-\frac{8}{17}$
$=\frac{15-8}{17}$
$=\frac{7}{17}$
Hence $\sin\text{A}-\cos\text{A}=\frac{7}{17}$
To find: $\sin\text{A}-\cos\text{A}$
Now we know $\tan\theta$ is defined as follows
$\tan\text{A}=\frac{\text{Perpendicular side oppsite to }\angle\text{A}}{\text{Base side adjacement to }\angle\text{A}}\ \dots(2)$
Now by comparing equation $(1)$ and $(2)$ We get, Perpendicular side opposite to $\angle\text{A}=15$ Base side adjacement to $\angle\text{A}=8$ Therefore triangle representing angle A is as shown below
Side AC is unknown and can be found using Pythagoras theorem Therefore, $AC^2 = AB^2 + BC^2$
Now by substituting the value of known sides from figure (a) We get,
$AC^2 = 15^2 + 8^2 = 225 + 64 = 289$
Now by taking square root on both sides We get, $\text{AC}=\sqrt{289}$ $=17$ Therefore Hypotenuse side AC = 17 …… (3)
Now we know, sin A is defined as follows
$\sin\text{A}=\frac{\text{Peependicular side opposite to}\angle\text{A}}{\text{Hypotenuse}}$ Therefore from figure (a) and equation (3) We get, $\sin\text{A}=\frac{\text{BC}}{\text{AC}}$ $=\frac{15}{17}$ $\sin\text{A}=\frac{15}{17}\ \dots(4)$
Now we know, cos A is defined as follows
$\cos\text{A}=\frac{\text{Base side adjacement to}\angle\text{A}}{\text{Hypotenuse}}$
Therefore from figure (a) and equation (3) We get, $\cos\text{A}=\frac{\text{AB}}{\text{AC}}$
$=\frac{8}{17}$ $\cos\text{A}=\frac{8}{17}\ \dots(5)$
Now we need to find the value of expression $\sin\text{A}-\cos\text{A}$
Therefore by substituting the value of sin A and cos A from equation $(4)$ and $(5)$ respectively,
we get, $\sin\text{A}-\cos\text{A}=\frac{15}{17}-\frac{8}{17}$
$=\frac{15-8}{17}$
$=\frac{7}{17}$
Hence $\sin\text{A}-\cos\text{A}=\frac{7}{17}$
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