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dlinn [17]
3 years ago
15

In ∆PQR, ∠P = 90degrees , PQ= 12 cm, PR=16cm. Find ∠Q

Mathematics
1 answer:
siniylev [52]3 years ago
5 0

Answer:

∠Q  = 53.13 degrees

Step-by-step explanation:

Given that ∆PQR, ∠P = 90degrees , this means that the triangle is a right angled triangle

Hence using the notation

SOH CAH TOA

where S is sine, C is cosine, T is tangent and O, A and H represents the size of the opposite, adjacent and hypotenuse sides

Considering  ∠Q and the given sides

PR=16cm is the opposite side,

PQ= 12 cm is the adjacent side hence we use TOA

Tan Q = 16/12 =

Q = Arc tan 16/12

= 53.13 degrees

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hoa [83]

Answer:

Part 1) x=16\ units

Part 2) x=23\ units

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Part 5) x=26.67\ in

Step-by-step explanation:

Part 1) we know that

If two triangles are similar

then

the ratio of their corresponding sides are equal

so

In this problem

\frac{12}{3}=\frac{x}{4}\\\\3x=12*4\\ \\x=48/3\\ \\x=16\ units

Part 2) we know that

If two triangles are similar

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the ratio of their corresponding sides are equal

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In this problem

\frac{36}{12}=\frac{39}{(x-10)}\\\\3(x-10)=39\\ \\x=(39/3)+10\\ \\x=23\ units

Part 3) we know that

If two triangles are similar

then

the ratio of their corresponding sides are equal

so

In this problem

\frac{AB}{CD}=\frac{AE}{ED}

substitute the values

\frac{10}{4}=\frac{2x+10}{x+3}

2.5(x+3)=2x+10

2.5x+7.5=2x+10

2.5x-2x=10-7.5

0.5x=2.5

x=5\ units

AE=2x+10=2*5+10=20\ units

Part 4) we know that

If two triangles are similar

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the ratio of their corresponding sides are equal

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In this problem

\frac{4}{5}=\frac{5}{x}\\ \\4x=5*5\\ \\ x=25/4\\ \\ x= 6.25\ units

Part 5) we know that

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10*10=100*10=1000*10=10000*10=100000*10=1000000*100=100,000,000

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ASAP!!! LOOK AT THE IMAGE FOR THE GRAPH AND ANSWER CHOICES!! PLEASE HELP URGENT!!
m_a_m_a [10]

Answer:

The answer is D.

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When given a graph with an intersection of two lines, first locate the y-intercept (the point where x = 0) of each line, and then find it's slope by evaluating its change in y over the change in x.

It is also possible to eliminate the incorrect systems using substitution when given a point of intersection.

For example, as the point of intersection is apparent on this graph, you can substitute these coordinates from the graph into each of the systems of equations given in the choice answers to verify it.

For it to be a solution, it must satisfy (make true) both equations of the system.

(a, b) : (x, y)

Given that (1,-2) is our point of intersection according to the graph.

Choice A is incorrect because (1, -2) → x + 4y = 3 → (1) + 4(-2) = 3 → 1 + (-8) = 3 →

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And (1, -2) → x + y = 2 → (1) + (-2) = 2 →

-1 ≠ 2.

It is already incorrect as one of the

equations in this system do not satisfy the coordinate.

Choice B is also incorrect because

(1, -2) → x + 4y = 2 → (1) + 4(-2) = 2 →

(1) + (-8) = 2 → 1 - 8 = 2 → -7 ≠ 2.

And (1, -2) → x + y = 3 → (1) + (-2) = 3 →

1 - 2 = 3 → -1 ≠ 3.

Lastly, Choice C is incorrect because

(1, -2) → 4x + y = 3 → 4(1) + (-2) = 3 →

4 + (-2) = 4 - 2 = 3 → 2 ≠ 3.

And (1, -2) → x - y = 2 → (1) - (-2) = 2 →

1 + 2 = 2 → 3 ≠ 2.

Therefore D is correct because it is the last answer remaining.

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4 + (-2) = 2 → 4 - 2 = 2 → <u>2 = 2</u>

So one equation fits the coordinate, but this cannot yet be verified as the working system unless both equations fit the point.

So (1, -2) → x - y = 3 → (1) - (-2) = 3 →

1 + 2 = 3 → <u>3 = 3</u>

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_____________________________

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Slope intercept form is the form y = mx + b where m is the slope( rise over run/change in y over change in x) , and b is the y-intercept (where x = 0).

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