approximate value of 3√997
Answers
Step-by-step explanation:
Given:
7tanθ=4
To find:
(7sinθ-3cosθ) / (7sinθ+3cosθ)
Solution:
We know that,
tanθ=\frac{perpendicular}{base}
base
perpendicular
We have,
7tanθ=4
tanθ=\frac{4}{7}=
7
4
So, we can say that perpendicular is 4units and base is 7units.
Now,
hypotenuse=\sqrt{(perpendicular)^2+(base)^2}hypotenuse=
(perpendicular)
2
+(base)
2
=\sqrt{4^2+7^2}=
4
2
+7
2
=\sqrt{16+49}=
16+49
=\sqrt{65}=
65
units
We know that,
sinθ=\frac{perpendicular}{hypotenuse}=
hypotenuse
perpendicular
=\frac{4}{\sqrt{65} }
65
4
And we know,
cosθ=\frac{base}{hypotenuse}=
hypotenuse
base
=\frac{7}{\sqrt{65} }
65
7
Now evaluating the numerator,
7sinθ-3cosθ
=7×\frac{4}{\sqrt{65} }
65
4
-4×\frac{7}{\sqrt{65} }
65
7
=\frac{28-21}{\sqrt{65} }
65
28−21
=\frac{7}{\sqrt{65} }=
65
7
Now, evaluating the denominator,
7sinθ+3cosθ
=7×\frac{4}{\sqrt{65} }
65
4
+4×\frac{7}{\sqrt{65} }
65
7
=\frac{28+21}{\sqrt{65} }
65
28+21
=\frac{49}{\sqrt{65} }=
65
49
So,
(7sinθ-3cosθ) / (7sinθ+3cosθ)
=\frac{\frac{7}{\sqrt{65} } }{\frac{49}{\sqrt{65} } }
65
49
65
7
=\frac{7}{\sqrt{65} }
65
7
×\frac{\sqrt{65} }{49}
49
65
=\frac{1}{7}=
7
1
Hence, the required value of (7sinθ-3cosθ) / (7sinθ+3cosθ) is \frac{1}{7}
7
1
.