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Reil [10]
2 years ago
14

In a rhombus MPKN with an obtuse angle K the diagonals intersect each other at point E. The measure of one of the angles of a ∆P

KE is equal 16°. Find the measures of all angles of ∆PKE and ΔPMN. Please help!
Mathematics
2 answers:
ololo11 [35]2 years ago
5 0
Angles PMN and MPN are also 16 because the triangles are isosceles.
The measure of the third angle (angle N) is 148 which also equal to the measure of angle K. The measure of angle PKE is 74 which half of the measure of angle N and angle K.<span />
Oduvanchick [21]2 years ago
4 0

Answer:

m∠KEP = 90°

m∠EKP = 74°

m∠EPK = 16°

m∠N PM= 16°

m∠MNP = 16

m∠M = 148°

Step-by-step explanation:

If you do RSM those are the answers

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CL 6-131. Solve each system using the method of your choice.
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System 1: The solution is (x, y) = (-4, 5)

System 2:  The solution is (x, y) = (\frac{11}{3}, -3)

<em><u>Solution:</u></em>

<em><u>Given system of equations are:</u></em>

2x + 3y = 7 ------ eqn 1

-3x - 5y = -13 --------- eqn 2

We can solve by elimination method

Multiply eqn 1 by 3

6x + 9y = 21 ------ eqn 3

Multiply eqn 2 by 2

-6x - 10y = -26 ------- eqn 4

Add eqn 3 and eqn 4

6x + 9y -6x - 10y = 21 - 26

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<h3><em><u>Second system of equation is:</u></em></h3>

8 - y = 3x ------ eqn 1

2y + 3x = 5 ----- eqn 2

We can solve by susbtitution method

From given,

y = 8 - 3x ----- eqn 3

Substitute eqn 3 in eqn 2

2(8 - 3x) + 3x = 5

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3x = 16 - 5

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x = \frac{11}{3}

Substitute the above value of x in eqn 3

y = 8 - 3x

y = 8 - 3 \times \frac{11}{3}\\\\y = 8 - 11\\\\y = -3

Thus the solution is (x, y) = (\frac{11}{3}, -3)

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