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dsp73
3 years ago
14

Lines m and n are parallel, as shown in the diagram below. What are the measures of angles A and B?

Mathematics
2 answers:
Dvinal [7]3 years ago
7 0

Answer:

a = 55° and b = 60°

Step-by-step explanation:

→ Remember 2 key points about angles

  • Angles in a triangle add up to 180°
  • Alternate angles are equal

→ Angle a is alternate to 55° so using the 2nd point,

a = 55°

→ Remember the fact that angle in a triangle add up to 180°

55°  + 65° + b = 180°

→ Collect the whole numbers

120° + b = 180°

→ Minus 120° from both sides to isolate b

b = 60°

baherus [9]3 years ago
3 0

Step-by-step explanation:

since m and n are parallel so the right side of b are equal 65degree. 180-55-65 to get B is 60 degree. After that, 180-60-65 will get the A is 55degree

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In a quadrilateral ABCD, angle A + angle D = 90°. Prove that AC^2 + BD^2 = BC^2 + AD^2​
olga2289 [7]

Answer:

AD^{2}+BC^{2}=AE^{2}+ED^{2}+BE^{2}+EC^{2}

AC^{2}+BD^{2}=AE^{2}+EC^{2}+BE^{2}+ED^{2}

Hence prove.

AC^{2}+BD^{2}=BC^{2}+AD^{2}

Step-by-step explanation:

Given:

∠A + ∠D = 90°

We are prove to that

AC^{2}+BD^{2}=BC^{2}+AD^{2}

Solution:

See required figure in attached file.

We know sum of the all angles of a triangle is 180°.

So, in triangle AED.

∠A + ∠E + ∠D = 180°

∠E + (∠A + ∠D) = 180°

Now, we substitute ∠A + ∠D = 90° in above equation.

∠E + 90° = 180°

∠E = 180° - 90°

∠E = 90°

Using Pythagoras Theorem for triangle ADE and triangle BEC.

AD^{2}=AE^{2}+ED^{2}

BC^{2}=BE^{2}+EC^{2}

Now, we add both above equations.

AD^{2}+BC^{2}=AE^{2}+ED^{2}+BE^{2}+EC^{2}--------(1)

Similarly, Using Pythagoras Theorem for triangle AEC and triangle BED.

AC^{2}=AE^{2}+EC^{2}

BD^{2}=BE^{2}+ED^{2}

Now, we add both above equations.

AC^{2}+BD^{2}=AE^{2}+EC^{2}+BE^{2}+ED^{2}--------(2)

We get From equation 1 and equation 2.

AC^{2}+BD^{2}=BC^{2}+AD^{2}

Hence prove,

AC^{2}+BD^{2}=BC^{2}+AD^{2}

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3 years ago
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