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Brilliant_brown [7]
3 years ago
11

Brochure that has the following problems with solutions.

Mathematics
1 answer:
DIA [1.3K]3 years ago
8 0

Answer:

P ≡ (- \frac{5}{4}, \frac{15}{4})

Step-by-step explanation:

The point P divides the line segment from A(-2,3) and B(1,6) in the ratio of 1 : 3.

So, AP : PB = 1 : 3.

Now, the coordinates of point P will be given by (\frac{1 \times 1 + 3 \times (- 2)}{1 + 3}, \frac{1 \times 6 + 3 \times 3}{1 + 3})

= (- \frac{5}{4}, \frac{15}{4}) (Answer)

Note: Let there are two points with known coordinates (x_{1},y_{1}) and (x_{2},y_{2}) and another a point having coordinates (h,k) divides the line joining the two above points internally in the ratio m : n, then (h,k) is given by

(h,k) ≡ (\frac{mx_{2} + nx_{1}}{m + n}, \frac{my_{2} + ny_{1}}{m + n})

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Zina [86]

Question:

If cos(θ) =-8/17 and sin(θ) is negative, then sin(θ) = ___  and tan(θ) =___.

Answer:

Sin\theta = \frac{-15}{17}

Tan\theta = \frac{15}{8}

Step-by-step explanation:

Given

cos(θ) =-8/17

Required

sin(θ) = __

tan(θ) =__

The first step is to determine the length of the third side

Given that

cos(\theta) = \frac{Adj}{Hyp}

Where Adj and Hyp represent Adjacent and Hypotenuse

cos(\theta) = \frac{-8}{17}

By comparison

Adj = -8\ and\ Hyp = 17

Using Pythagoras

Hyp^2 = Adj^2 + Opp^2

By Substitution

17^2 = (-8)^2 + Opp^2

289 = 64 + Opp^2

Subtract 64 from both sides

289 - 64 = 64 - 64 + Opp^2

225 = Opp^2

Take square roots of both sides

\sqrt{225} = \sqrt{Opp^2}

\sqrt{225} = Opp

15 = Opp

Opp = 15

The question says that sin(θ) is negative; This implies that θ is in the third quadrant and as such

Opp = -15

From trigonometry

Sin\theta = \frac{Opp}{Hyp}

Sin\theta = \frac{-15}{17}

Also from trigonometry

Tan\theta = Sin\theta / Cos\theta

Tan\theta = \frac{-15}{17} / \frac{-8}{17}

Tan\theta = \frac{-15}{17} * \frac{-17}{8}

Tan\theta = \frac{-15 * -17}{17  * 8}

Tan\theta = \frac{15 * 17}{17  * 8}

Tan\theta = \frac{15}{8}

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