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Mathematics

1 answer:

Scilla [17]3 years ago

3 0

The formula of a midpoint:

$\left(\dfrac{x_1+x_2}{2};\ \dfrac{y_1+y_2}{2}\right)$

D is midpoint of TR.

We have T(-2a, 0) and R(-2b, 2c). Substitute:

$x=\dfrac{-2a+(-2b)}{2}=\dfrac{-2a-2b}{2}=\dfrac{2(-a-b)}{2}=-a-b\\\\y=\dfrac{0+2c}{2}=\dfrac{2C}{2}=c$

<h3>Answer: b. (-a - b, c)</h3>

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The point (5, −2) is on the terminal ray of angle θ, which is in standard position. Without evaluating, explain how you would fi

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1. Check the picture attached.

let P, M, N be the points :P(5, -2), M(5, 0), N(0, -2)

ABP is the angle with measure ∅, where BA is the initial ray and BP the terminal ray.

2. m(PNM)=∅ as can be easily checked

In triangle PNM, NP=5, PM=2 and NM= $\sqrt{ 5^{2} + 2^{2} }= \sqrt{25+4}= \sqrt{29}$ units

3. let opp=opposite side, adj=adjecent side, hyp=hypothenuse

sin ∅ = opp/hyp= $\frac{2}{\sqrt{29}}$

cos ∅ = adj/hyp= $\frac{5}{\sqrt{29}}$

tan ∅ = opp/adj=2/5

cot ∅ = adj/opp=5/2

sec ∅ = hyp/adj= $\frac{\sqrt{29}}{5}$

csc ∅ = hyp/opp= $\frac{\sqrt{29}}{2}$