Given sec A= and that angle A is in Quadrant I, find the exact value of csc A in
simplest radical form using a rational denominator.
Need help :(
Answers
The required value is [tex]\csc A=\dfrac{5}{4}[/tex].
Important information:
[tex]\sec A=\dfrac{5}{3}[/tex]Angle A is in Quadrant I.We need to find the exact value of csc A.
Trigonometric Ratios:Formulae used: If angle A is in Quadrant I.
[tex]\cos A=\dfrac{1}{\sec A}[/tex]
[tex]\sin A=\sqrt{1-\cos^2 A}[/tex]
[tex]\csc A=\dfrac{1}{\sin A}[/tex]
Using these formulae, we get
[tex]\cos A=\dfrac{3}{5}[/tex]
And,
[tex]\sin A=\sqrt{1-\left(\dfrac{3}{5}\right)^2}[/tex]
[tex]\sin A=\sqrt{1-\dfrac{9}{25}}[/tex]
[tex]\sin A=\sqrt{\dfrac{16}{25}}[/tex]
[tex]\sin A=\dfrac{4}{5}[/tex]
Now,
[tex]\csc A=\dfrac{1}{\sin A}[/tex]
[tex]\csc A=\dfrac{1}{\frac{4}{5}}[/tex]
[tex]\csc A=\dfrac{5}{4}[/tex]
Thus, the required value is [tex]\csc A=\dfrac{5}{4}[/tex].
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The required value of cscA is 1.25
Given,
[tex]secA=\frac{5}{3}[/tex]
Trigonometrical ratios:
In the first quadrant, all ratios are positives.
[tex]sin A=\frac{P}{H}[/tex]
And the value of hypotenuse is,
[tex]P=\sqrt{H^2-B^2} \\=\sqrt{5^2-3^2} \\=\sqrt{16}\\ P=4[/tex]
So, the sine A is,
[tex]sinA=\frac{P}{H}\\ =\frac{4}{5}[/tex]
So, the rational denominator is,
[tex]cscA=\frac{H}{P}\\ =\frac{5}{4}\\=1.25[/tex]
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[tex]\textsf{Hey there!}[/tex]
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Answers
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Step-by-step explanation:
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What is proportion ?
A proportion is an equation based on the equality of two ratios.
Here it is given that :
Alaina is making a birdseed mix with 2 1/3 lb sunflower seeds, 5/6 lb cracked corn, and 1/6 lb peanuts.
adding everything to make birdseed mix = 2 1/3 + 5/6 + 1/6 =10/3 lb
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for making 10/3 lb birdseed mix = 5/6 lb cracked corn is used
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= 15/12 = 5/4 = 1 ¹/₄ lb cracked corn will be used.
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