11.01.2020

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°

m∠DBC =
m∠DCB =
m∠CDB =
m∠ACD =
m∠ADC =

. 8

Faq

Mathematics
Step-by-step answer
P Answered by Specialist

Given :

ZC = 90°CD is the altitude to AB.\angleA = 65°.

To find :

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

Solution :

In Right angle △ABC,

we have,

=> ACB = 90°

=> \angleCAB = 65°.

So,

=> \angleACB + \angleCAB+\angleZCBA = 180° (By angle sum Property.)

=> 90° + 65° + \angleCBA = 180°

=> 155° +\angleCBA = 180°

=> \angleCBA = 180° - 155°

=> \angleCBA = 25°.

In △CDB,

=> CD is the altitude to AB.

So,

=> \angle CDB = 90°

=> \angleCBD = \angleCBA = 25°.

So,

=> \angleCBD + \angleDCB = 180° (Angle sum Property.)

=> 90° +25° + \angleDCB = 180°

=> 115° + \angleDCB = 180°

=> \angleDCB = 180° - 115°

=> \angleDCB = 65°.

Now, in △ADC,

=> CD is the altitude to AB.

So,

=> \angleADC = 90°

=>\angle CAD =\angle CAB = 65°.

So,

=> \angleADC + \angleCAD +\angleDCA = 180° (Angle sum Property.)

=> 90° + 65° + \angleDCA = 180°

=> 155° +\angleDCA = 180°

=> \angleDCA = 180° - 155°

=> \angleDCA = 25°

Hence, we get,

\angleDCA = 25°\angleDCB = 65°\angleCDB = 90°\angleACD = 25°\angleADC = 90°.
Mathematics
Step-by-step answer
P Answered by Specialist

Given :

ZC = 90°CD is the altitude to AB.\angleA = 65°.

To find :

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

Solution :

In Right angle △ABC,

we have,

=> ACB = 90°

=> \angleCAB = 65°.

So,

=> \angleACB + \angleCAB+\angleZCBA = 180° (By angle sum Property.)

=> 90° + 65° + \angleCBA = 180°

=> 155° +\angleCBA = 180°

=> \angleCBA = 180° - 155°

=> \angleCBA = 25°.

In △CDB,

=> CD is the altitude to AB.

So,

=> \angle CDB = 90°

=> \angleCBD = \angleCBA = 25°.

So,

=> \angleCBD + \angleDCB = 180° (Angle sum Property.)

=> 90° +25° + \angleDCB = 180°

=> 115° + \angleDCB = 180°

=> \angleDCB = 180° - 115°

=> \angleDCB = 65°.

Now, in △ADC,

=> CD is the altitude to AB.

So,

=> \angleADC = 90°

=>\angle CAD =\angle CAB = 65°.

So,

=> \angleADC + \angleCAD +\angleDCA = 180° (Angle sum Property.)

=> 90° + 65° + \angleDCA = 180°

=> 155° +\angleDCA = 180°

=> \angleDCA = 180° - 155°

=> \angleDCA = 25°

Hence, we get,

\angleDCA = 25°\angleDCB = 65°\angleCDB = 90°\angleACD = 25°\angleADC = 90°.
Mathematics
Step-by-step answer
P Answered by PhD

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°.

Mathematics
Step-by-step answer
P Answered by PhD

1. ∆CBD:      ∠B = 70°; ∠BCD = 20°; ∠ BDC = 90°

2. ∆CDA: ∠ACD = 70°;      ∠A = 20°; ∠ ADC = 90°

Step-by-step explanation:

1. ∆DBC

In ∆ABC

 ∠A + ∠B + ∠C = 180°

20° + ∠B + 90 ° = 180°

         ∠B + 110 ° = 180°

      ∠DBC = ∠B =  70°

In ∆CBD

                      ∠BDC =  90°

∠B + ∠BCD + ∠CBD = 180°

  70° + ∠BCD + 90 ° = 180°

           ∠BCD + 160° = 180°

                        BCD =   20°

2. ∆CAD

∠A + ∠ACD + ∠ADC = 180°

   20° + ∠ACD + 90° = 180°

            ∠ACD + 110° = 180°

                      ∠ACD =   70°


In triangle ΔABC, ∠C is a right angle and CD is the altitude to AB . Find the measures of the angles
Mathematics
Step-by-step answer
P Answered by PhD

1. ∆CBD:      ∠B = 70°; ∠BCD = 20°; ∠ BDC = 90°

2. ∆CDA: ∠ACD = 70°;      ∠A = 20°; ∠ ADC = 90°

Step-by-step explanation:

1. ∆DBC

In ∆ABC

 ∠A + ∠B + ∠C = 180°

20° + ∠B + 90 ° = 180°

         ∠B + 110 ° = 180°

      ∠DBC = ∠B =  70°

In ∆CBD

                      ∠BDC =  90°

∠B + ∠BCD + ∠CBD = 180°

  70° + ∠BCD + 90 ° = 180°

           ∠BCD + 160° = 180°

                        BCD =   20°

2. ∆CAD

∠A + ∠ACD + ∠ADC = 180°

   20° + ∠ACD + 90° = 180°

            ∠ACD + 110° = 180°

                      ∠ACD =   70°


In triangle ΔABC, ∠C is a right angle and CD is the altitude to AB . Find the measures of the angles
Mathematics
Step-by-step answer
P Answered by PhD

Because f+e=c

Therefore ,a^2+b^2=c^2

Step-by-step explanation:

Given A right triangle ABC as shown in figure  where CD is an altitude of the triangle.

To prove that a^2+b^2=c^2

Proofe: Given \triangle ABC\; and\; \triangle CBD both are right triangle and both triangles have common angle B si same.

Therefore , two angles of two triangles are equal .

Hence, \triangle ABC \sim\triangle CBD by using AA similarity.

Similarity property: when two triangles are similar then their corresponding angles are equal and their corresponding side are in equal proportion.\frac{a}{f} =\frac{c}{a}

Similarly , \triangle ABC \sim \triangle ACD by AA similarity property . Because both triangles are right triangles therefore, one angle of both triangles is equal to 90 degree and both triangles have one common angle A is same .

\therefore\frac{b}{c} =\frac{e}{b}

The corresponding parts of two similar triangles are in equal proportion  therefore , two proportion can be rewrite as

a^2=cf (I equation)

and  b^2=ce (II equation)

Adding b^2 to  both sides of firs equation

a^2+b^2=cf+b^2

Because b^2=ce and ce can be substituted into the right side of equation wevcan write as

a^2+b^2=ce+cf

Applying the converse of distributive property we can write

a^2+b^2=c(f+e)

Distributive property: a.(b+c)= a.c+a.b

a^2+b^2=c^2

Because f+e=c^2

Hence proved.


Consider the diagram and the paragraph proof below. given:  right △abc as shown where cd is an altit
Mathematics
Step-by-step answer
P Answered by PhD
Given: Right △ABC as shown where CD is an altitude of the triangle. We  prove that a^2 + b^2 = c^2

Because △ABC and △CBD both have a right angle, and the same angle B is in both triangles, the triangles must be similar by AA.

Likewise, △ABC and △ACD both have a right angle, and the same angle A is in both triangles, so they also must be similar by AA.

The proportions \frac{c}{a} and \frac{a}{f} are true because they are ratios of corresponding parts of similar triangles.

The two proportions can be rewritten as a^2 = cf and b^2 = ce.

Adding b^2 to both sides of first equation, a^2 = cf, results in the equation a^2 + b^2 = cf + b^2.

Because b^2 and ce are equal, ce can be substituted into the right side of the equation for b^2, resulting in the equation a^2 + b^2 = cf + ce.

Applying the converse of the distributive property results in the equation a^2 + b^2 = c(f + e).

The last sentence of the proof is

Because f + e = c, a^2 + b^2 = c^2.
Mathematics
Step-by-step answer
P Answered by Specialist

Given: Right △ABC as shown where CD is an altitude of the triangle. We prove that a^2+b^2=c^2


Because △ABC and △CBD both have a right angle, and the same angle B is in both triangles, the triangles must be similar by AA.


Likewise, △ABC and △ACD both have a right angle, and the same angle A is in both triangles, so they also must be similar by AA.


The proportions c\a and a\f are true because they are ratios of corresponding parts of similar triangles.


The two proportions can be rewritten as a^2=cf and b^2=ce



Adding b^2 to both sides of first equation, a^2=cf , results in the equation a^2 + b^2 = cf+ b^2 .



Because b^2 and ce are equal, ce can be substituted into the right side of the equation for b^2 , resulting in the equation a^2 + b^2 = cf + ce .



Applying the converse of the distributive property results in the equation .

a^2 + b^2 = c (f + e)


The last sentence of the proof is

Because f + e = c, a^2 + b^2 = c^2 .

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