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Answer for a lot of points!
In triangle △ABC, ∠C is a right angle and CD is the altitude to AB . Find the angles in △CBD and △CAD if m∠A = 65°
m∠DBC =
m∠DCB =
m∠CDB =
m∠ACD =
m∠ADC =

Respuesta :

Given :

  • ZC = 90°

  • CD is the altitude to AB.

  • [tex]\angle[/tex]A = 65°.

To find :

  • the angles in △CBD and △CAD if m∠A = 65°

Solution :

In Right angle △ABC,

we have,

=> ACB = 90°

=> [tex]\angle[/tex]CAB = 65°.

So,

=> [tex]\angle[/tex]ACB + [tex]\angle[/tex]CAB+[tex]\angle[/tex]ZCBA = 180° (By angle sum Property.)

=> 90° + 65° + [tex]\angle[/tex]CBA = 180°

=> 155° +[tex]\angle[/tex]CBA = 180°

=> [tex]\angle[/tex]CBA = 180° - 155°

=> [tex]\angle[/tex]CBA = 25°.

In △CDB,

=> CD is the altitude to AB.

So,

=> [tex]\angle[/tex] CDB = 90°

=> [tex]\angle[/tex]CBD = [tex]\angle[/tex]CBA = 25°.

So,

=> [tex]\angle[/tex]CBD + [tex]\angle[/tex]DCB = 180° (Angle sum Property.)

=> 90° +25° + [tex]\angle[/tex]DCB = 180°

=> 115° + [tex]\angle[/tex]DCB = 180°

=> [tex]\angle[/tex]DCB = 180° - 115°

=> [tex]\angle[/tex]DCB = 65°.

Now, in △ADC,

=> CD is the altitude to AB.

So,

=> [tex]\angle[/tex]ADC = 90°

=>[tex]\angle[/tex] CAD =[tex]\angle[/tex] CAB = 65°.

So,

=> [tex]\angle[/tex]ADC + [tex]\angle[/tex]CAD +[tex]\angle[/tex]DCA = 180° (Angle sum Property.)

=> 90° + 65° + [tex]\angle[/tex]DCA = 180°

=> 155° +[tex]\angle[/tex]DCA = 180°

=> [tex]\angle[/tex]DCA = 180° - 155°

=> [tex]\angle[/tex]DCA = 25°

Hence, we get,

  • [tex]\angle[/tex]DCA = 25°
  • [tex]\angle[/tex]DCB = 65°
  • [tex]\angle[/tex]CDB = 90°
  • [tex]\angle[/tex]ACD = 25°
  • [tex]\angle[/tex]ADC = 90°.

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