Respuesta :
In triangle ΔABC,
<C= 90° ( Given angle is is a right angle).
m∠A = 65° (Also given).
Sum of angles of a triangle is 180°.
We can set an equation for angles A, B and C.
<A+<B+<C = 180°.
Plugging values of <A and <C in above eqaution.
90° + <B + 65° = 180°
<B + 155 = 180.
Subtracting 155 from both sides.
<B +155 - 155 = 180-155.
<B = 25°.
Therefore, <B = 25°.
Now, in triangle ΔCBD.
<D = 90°. (Give CD is perpendicular to AB. A perpendicular line subtands an angle 90 degrees).
< B= 25° ( We found above).
Now, sum of angles of triangle ΔCBD is also 180 degree.
We can setup another equation,
<B + <D + < BCD = 180 °.
Plugging values of B and D in above equation.
25 + 90 + <BCD = 180.
115 + < BCD = 180.
Subtracting 115 from both sides.
115 + < BCD -115 = 180-115.
<BCD = 65°.
Now, in triangle ΔCAD.
<D = 90°.
<A = 65°
We need to find <ACD.
Now, sum of angles of triangle ΔCAD is also 180 degree.
We can setup another equation,
<A + <D + < ACD = 180 °.
65 + 90 + <ACD = 180.
155 + <ACD = 180.
Subtracting 155 from both sides.
<ACD +155 - 155 = 180-155.
<ACD = 25°.
Therefore, <ACD= 25°, <BCD = 65°, <D=90°, <B= 25°.
Answer:
In triangle CBD
[tex]\angle CBD=25^{\circ}, \angle CDB=90^{\circ}, \angle BCD=65^{\circ}[/tex]
In triangle CAD
[tex]\angle CAD=65^{\circ}, \angle CDA=90^{\circ}, \angle ACD=25^{\circ}[/tex]
Step-by-step explanation:
In triangle ABC
[tex]\angle A=65^{\circ}, \angle C=90^{\circ}[/tex]
We know that sum of angles of a triangle =180 degrees
[tex]\angle A+\angle B+\angle C=180^{\circ}[/tex]
[tex]65+90+\angle B=180[/tex]
By using substituting property of equality
[tex]155+\angle B=180[/tex]
[tex]\angle B=180-155[/tex]
By subtraction property of equality
[tex]\angle B= 25^{\circ}[/tex]
In triangle CBD
[tex]\angle CBD=25^{\circ}, \angle CDB=90^{\circ} [/tex]
Because CD is perpendicular to AB.
[tex]\angle CBD+\angle CDB+\angle BCD=180^{\circ}[/tex]
[tex]25+90+\angle BCD=180[/tex]
By substitution property
[tex]115+\angle BCD=180[/tex]
[tex]\angle BCD=180-115[/tex]
By subtraction property of equality
[tex]\angle BCD=65^{\circ}[/tex]
In triangle CAD
[tex]\angle CAD=65^{\circ}, \angle CDA=90^{\circ}[/tex]
Because CD is perpendicular to AB.
[tex]\angle CAD+\angle CDA+\angle ACD=180^{\circ}[/tex]
[tex]65+90+\angle ACD=180[/tex]
By using substitution property
[tex]155+\angle ACD=180[/tex]
[tex]\angle ACD=180-155[/tex]
By subtraction property of equality
[tex]\angle ACD=25^{\circ}[/tex].
