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3 years ago

In a right triangle, the acute angles measure 2x+50 and 3x degrees

2 answers:

8 0

Triangle int. angles add up to 180 degrees.
In a right triangle, there are 2 acute angles and 1 right angle.

Here are the angles given:
a1) 90°
a2) (2x + 50)°
a3) (3x)°

Use the equation of the sum of interior angles on a triangle and solve for x.

180 = a1 + a2 + a3
180 = 90 + (3x) + (2x + 50)
180 = 90 + 5x + 50
180 = 140 + 5x
40 = 5x
/5 /5
8 = x

Therefore x = 8. Now we can put these values into the acute angles to get the measurement of all angles.

a1 = 90

a2 = 2x + 50
= 2(8) + 50
= 16 + 50
= 66

a3 = 3x
= 3(8)
= 24

Therefore, the measurement of all the angles in the triangle is 90°, 66°, and 24°.

Proof:

180 = a1 + a2 + a3
180 = 90 + 66 + 24
180 = 156 + 24
180 = 180

So it is true. Hope I helped.

xenn [34]3 years ago

6 0

2x+50 + 3x= 90 degrees
5x +50 = 90
5x =40
x=8

Angle1= 2(8) +50= 66
Angle2= 3(8)=24

66+24+90= 180

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Perform the indicated row operations, then write the new matrix.

Studentka2010 [4]

The matrix is not properly formatted.

However, I'm able to rearrange the question as:

$\left[\begin{array}{ccc}1&1&1|-1\\-2&3&5|3\\3&2&4|1\end{array}\right]$

Operations:

$2R_1 + R_2 ->R_2$

$-3R_1 +R_3 ->R_3$

Please note that the above may not reflect the original question. However, you should be able to implement my steps in your question.

Answer:

$\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]$

Step-by-step explanation:

The first operation:

$2R_1 + R_2 ->R_2$

This means that the new second row (R2) is derived by:

Multiplying the first row (R1) by 2; add this to the second row

The row 1 elements are:

$\left[\begin{array}{ccc}1&1&1|-1\end{array}\right]$

Multiply by 2

$2 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}2&2&2|-2\end{array}\right]$

Add to row 2 elements are: $\left[\begin{array}{ccc}-2&3&5|3\end{array}\right]$

$\left[\begin{array}{ccc}2&2&2|-2\end{array}\right] + \left[\begin{array}{ccc}-2&3&5|3\end{array}\right]$

$\left[\begin{array}{ccc}0&5&7|1\end{array}\right]$

The second operation:

$-3R_1 +R_3 ->R_3$

This means that the new third row (R3) is derived by:

Multiplying the first row (R1) by -3; add this to the third row

The row 1 elements are:

$\left[\begin{array}{ccc}1&1&1|-1\end{array}\right]$

Multiply by -3

$-3 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right]$

Add to row 2 elements are: $\left[\begin{array}{ccc}3&2&4|1\end{array}\right]$

$\left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right] + \left[\begin{array}{ccc}3&2&4|1\end{array}\right]$

$\left[\begin{array}{ccc}0&-1&1|4\end{array}\right]$

Hence, the new matrix is:

$\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]$

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