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2 years ago

Given cos 0 = 11/6 and angle 0 is in Quadrant I, what is the exact value of sin 0 in

Mathematics

1 answer:

user100 [1]2 years ago

8 0

1 = sin²0 + cos²0

sin²0 = 1 - cos²0

sin²0 = 1 - 11/36

sin²0 = 25/36

sin 0 = 5/6 or -5/6

In the first quadrant, the values for sin, cos and tan are positive.

sin0 = 5/6

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Given that cot θ = 1/√5, what is the value of (sec²θ - cosec²θ)/(sec²θ + cosec²θ) ?

Bogdan [553]

Step-by-step explanation:

$\mathsf{Given :\;\dfrac{{sec}^2\theta - co{sec}^2\theta}{{sec}^2\theta + co{sec}^2\theta}}$

$\bigstar\;\;\textsf{We know that : \large\boxed{\mathsf{{sec}\theta = \dfrac{1}{cos\theta}}}}$

$\bigstar\;\;\textsf{We know that : \large\boxed{\mathsf{co{sec}\theta = \dfrac{1}{sin\theta}}}}$

$\mathsf{\implies \dfrac{\dfrac{1}{cos^2\theta} - \dfrac{1}{sin^2\theta}}{\dfrac{1}{cos^2\theta} + \dfrac{1}{sin^2\theta}}}$

$\mathsf{\implies \dfrac{\dfrac{sin^2\theta - cos^2\theta}{sin^2\theta.cos^2\theta}}{\dfrac{sin^2\theta + cos^2\theta}{sin^2\theta.cos^2\theta}}}$

$\mathsf{\implies \dfrac{sin^2\theta - cos^2\theta}{sin^2\theta + cos^2\theta}}$

Taking sin²θ common in both numerator & denominator, We get :

$\mathsf{\implies \dfrac{sin^2\theta\left(1 - \dfrac{cos^2\theta}{sin^2\theta}\right)}{sin^2\theta\left(1 + \dfrac{cos^2\theta}{sin^2\theta}\right)}}$

$\bigstar\;\;\textsf{We know that : \large\boxed{\mathsf{cot\theta = \dfrac{cos\theta}{sin\theta}}}}$

$\mathsf{\implies \dfrac{1 -cot^2\theta}{1 + cot^2\theta}}$

$\mathsf{Given :\;cot\theta = \dfrac{1}{\sqrt{5}}}$

$\mathsf{\implies \dfrac{1 - \left(\dfrac{1}{\sqrt{5}}\right)^2}{1 + \left(\dfrac{1}{\sqrt{5}}\right)^2}}$

$\mathsf{\implies \dfrac{1 - \dfrac{1}{5}}{1 + \dfrac{1}{5}}}$

$\mathsf{\implies \dfrac{\dfrac{5 - 1}{5}}{\dfrac{5 + 1}{5}}}$

$\mathsf{\implies \dfrac{5 - 1}{5 + 1}}$

$\mathsf{\implies \dfrac{4}{6}}$

$\mathsf{\implies \dfrac{2}{3}}$

Hence, option (a) 2/3 is your correct answer.

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2 years ago

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SpyIntel [72]

Answer:

Step-by-step explanation:

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3 years ago

Click an item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the corre

GaryK [48]

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2 years ago

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kolezko [41]

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3 years ago

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Runner A is initially 6.0 km west of a flagpole and is running with a constant velocity of 3.0 km/h due east. Runner B is initia

ArbitrLikvidat [17]

Answer:

The distance of the two runners from the flagpole is 0.0882 km

Step-by-step explanation:

We are going to use the minus sign when the runner is on the west and the plus sign when the runner is on the east.

Then, Taking into account the runner A is 6 km west and is running with a constant velocity of 3 Km/h east, the distance of the runner A from the flagpole is given by the following equation:

Xa = -6 Km + (3 Km/h)* t

Where Xa is the position of the runner A from the flagpole and t is the time in hours.

At the same way the distance of the runner B, Xb, from the flagpole is given by the following equation:

Xb = 7.4 Km - (3.8 Km/h)*t

Then, the two runners cross their path when Xa is equal to Xb, so if we solve this equation for t, we get:

Xa = Xb

-6 + (3*t) = 7.4 - (3.8*t)

(3.8*t) + (3*t) = 7.4 + 6

(6.8*t) = 13.4

t = 13.4/6.8

t = 1.9706

Therefore, at time t equal to 1.9706 hours, both runners cross their path. The distance of the two runners from the flagpole can be calculated replacing the value of t in equation for Xa or in equation for Xb as:

Xa = -6 Km + (3 Km/h)* t

Xa = -6 Km + (3 Km/h)*(1.9706 h)

Xa = -0.0882 Km

That means that both runners are 0.0882 Km west of a flagpole.

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