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1 year ago

Use the information given in the diagrams to show that

Mathematics

1 answer:

Vlad1618 [11]1 year ago

7 0

Answer:

d = k·sin(2θ)·sin(α)/(sin(θ)·sin(β))

Step-by-step explanation:

The Law of Sines tells us that sides of a triangle are proportional to the sine of the opposite angle. This can be used along with a trig identity to demonstrate the required relation.

<h3>top triangle</h3>

The law of sines applied to the top triangle is ...

BC/sin(A) = AC/sin(θ)

Triangle ABC is isosceles, so the base angles at B and C are congruent. Then the angle at vertex A is ...

∠A = 180° -θ -θ = 180° -2θ

A trig identity tells us the sine of an angle is equal to the sine of its supplement. That means the sine of angle A is ...

sin(A) = sin(180° -2θ) = sin(2θ)

and our above Law of Sines equation tells us ...

BC = sin(A)/sin(θ)·AC = k·sin(2θ)/sin(θ)

<h3>bottom triangle</h3>

The law of sines applied to the bottom triangle is ...

DC/sin(B) = BC/sin(D)

d/sin(α) = BC/sin(β)

Multiplying by sin(α) we have ...

d = BC·sin(α)/sin(β)

Using our expression for BC gives the desired relation:

d = k·sin(2θ)·sin(α)/(sin(θ)·sin(β))

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mel-nik [20]

Option D. D has the matrix of constants [[12], [11], [4]].

Step-by-step explanation:

Step 1:

With the given equations, we can form matrices to represent them.

The coefficients of x, y, and z form a matrix of order 3 ×3, the variables x, y, and z form a matrix of order 1 ×3 and the constants form a matrix of order 1 ×3.

Step 2:

The linear system A is represented as

$\left[\begin{array}{ccc}12&1&1\\1&-11&0\\1&-1&4\end{array}\right]$ $\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$ $= \left[\begin{array}{ccc}26\\17\\23\end{array}\right]$ .

Step 3:

The linear system B is represented as

$\left[\begin{array}{ccc}4&1&1\\1&-11&0\\1&-1&12\end{array}\right]$ $\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$ $= \left[\begin{array}{ccc}23\\17\\26\end{array}\right]$ .

Step 4:

The linear system C is represented as

$\left[\begin{array}{ccc}1&1&1\\1&-1&0\\1&-1&1\end{array}\right]$ $\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$ $= \left[\begin{array}{ccc}4\\11\\12\end{array}\right]$ .

Step 5:

The linear system D is represented as

$\left[\begin{array}{ccc}1&1&1\\1&-1&0\\1&-1&1\end{array}\right]$ $\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$ $= \left[\begin{array}{ccc}12\\11\\4\end{array}\right]$ .

Step 6:

Of the four options, the linear system D has the matrix of constants [[12], [11], [4]]. So the answer is option D. D.

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Step-by-step explanation:

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