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

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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°.

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The given information can be used to determine the measures of the required angles, so that:

i. the measure of all angles in ΔCBD are:

m<BCD = [tex]65^{o}[/tex]

m<CDB = [tex]90^{o}[/tex]

m<B = [tex]25^{o}[/tex]

ii. the measure of angles in ΔCAD are:

m<A = 65°

m<ACD = [tex]25^{o}[/tex]

m<ADC = [tex]90^{o}[/tex]

From ΔABC, given that m∠A = 65° and <C = [tex]90^{o}[/tex] then;

m<A + m<B + m<C = [tex]180^{o}[/tex]

65° + m<B + [tex]90^{o}[/tex] = [tex]180^{o}[/tex]

m<B = [tex]180^{o}[/tex] - [tex]155^{o}[/tex]

       = [tex]25^{o}[/tex]

m<B = [tex]25^{o}[/tex]

a. CD is perpendicular to AB, so that:

m<ADC = m<BDC = [tex]90^{o}[/tex]

Thus,

m<A + m<ACD + m<ADC = [tex]180^{o}[/tex]

65° + m<ACD + [tex]90^{o}[/tex]  = [tex]180^{o}[/tex]

m<ADC = [tex]180^{o}[/tex] - [tex]155^{o}[/tex]

        = [tex]25^{o}[/tex]

m<ACD = [tex]25^{o}[/tex]

Also,

m<BCD + m<CDB + m<B = [tex]180^{o}[/tex]

m<BCD + [tex]90^{o}[/tex] + [tex]25^{o}[/tex] = [tex]180^{o}[/tex]

m<BCD = [tex]180^{o}[/tex] - [tex]115^{o}[/tex]

             = [tex]65^{o}[/tex]

m<BCD = [tex]65^{o}[/tex]

Therefore;

i. the measure of all angles in ΔCBD are:

m<BCD = [tex]65^{o}[/tex]

m<CDB = [tex]90^{o}[/tex]

m<B = [tex]25^{o}[/tex]

ii. the measure of angles in ΔCAD are:

m<A = 65°

m<ACD = [tex]25^{o}[/tex]

m<ADC = [tex]90^{o}[/tex]

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