Question

In triangle ΔABC, ∠C is a right angle and CD is the height to AB Find the angles in ΔCBD and ΔCAD if:m∠A = 65°

Answer 1

In triangle CBD

`\angle CBD=25^(\circ), \angle CDB=90^(\circ), \angle BCD=65^(\circ)`

In triangle CAD

`\angle CAD=65^(\circ), \angle CDA=90^(\circ), \angle ACD=25^(\circ)`

Explanation:

In triangle ABC

`\angle A=65^(\circ), \angle C=90^(\circ)`

We know that sum of angles of a triangle =180 degrees

`\angle A+\angle B+\angle C=180^(\circ)`

`65+90+\angle B=180`

By using substituting property of equality

`155+\angle B=180`

`\angle B=180-155`

By subtraction property of equality

`\angle B= 25^(\circ)`

In triangle CBD

`\angle CBD=25^(\circ), \angle CDB=90^(\circ)`

Because CD is perpendicular to AB.

`\angle CBD+\angle CDB+\angle BCD=180^(\circ)`

`25+90+\angle BCD=180`

By substitution property

`115+\angle BCD=180`

`\angle BCD=180-115`

By subtraction property of equality

`\angle BCD=65^(\circ)`

In triangle CAD

`\angle CAD=65^(\circ), \angle CDA=90^(\circ)`

Because CD is perpendicular to AB.

`\angle CAD+\angle CDA+\angle ACD=180^(\circ)`

`65+90+\angle ACD=180`

By using substitution property

`155+\angle ACD=180`

`\angle ACD=180-155`

By subtraction property of equality

`\angle ACD=25^(\circ)`.

Answer 2

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°.