Answer:
Step-by-step explanation:
In triangle ΔABC,
<C=90° (Given angle is a right angle).
m∠A=65° (Also given).
The sum of angles of a triangle is 180°.
We can set an equation for angles A, B and C.
<A+<B+<C=180°.
Now we are plugging values of <A and <C in the equation above.
90°+<B+65°=180°
<B+155=180.
Subtract 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 90° angle.)
<B=25° (We found above).
Now, the sum of the angles of triangle ΔCBD is also 180°.
We can set up another equation,
<B + <D + < BCD = 180 °.
Plugging values of B and D in the equation above.
25+90+<BCD=180.
115+<BCD=180.
Subtract 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, the sum of the angles of triangle ΔCAD is also 180 degrees.
We can set up another equation,
<A+<D+<ACD=180°.
65+90+<ACD=180.
155+<ACD=180.
Subtract 155 from both sides.
<ACD+155-155=180-155.
<ACD=25°.
Therefore, <ACD=25°, <BCD=65°, <D=90°, <B=25°.