Given :
ZC = 90°CD is the altitude to AB.A = 65°.To find :
the angles in △CBD and △CAD if m∠A = 65°Solution :
In Right angle △ABC,
we have,
=> ACB = 90°
=> CAB = 65°.
So,
=> ACB + CAB+ZCBA = 180° (By angle sum Property.)
=> 90° + 65° + CBA = 180°
=> 155° +CBA = 180°
=> CBA = 180° - 155°
=> CBA = 25°.
In △CDB,
=> CD is the altitude to AB.
So,
=> CDB = 90°
=> CBD = CBA = 25°.
So,
=> CBD + DCB = 180° (Angle sum Property.)
=> 90° +25° + DCB = 180°
=> 115° + DCB = 180°
=> DCB = 180° - 115°
=> DCB = 65°.
Now, in △ADC,
=> CD is the altitude to AB.
So,
=> ADC = 90°
=> CAD = CAB = 65°.
So,
=> ADC + CAD +DCA = 180° (Angle sum Property.)
=> 90° + 65° + DCA = 180°
=> 155° +DCA = 180°
=> DCA = 180° - 155°
=> DCA = 25°
Hence, we get,
DCA = 25°DCB = 65°CDB = 90°ACD = 25°ADC = 90°.Given :
ZC = 90°CD is the altitude to AB.A = 65°.To find :
the angles in △CBD and △CAD if m∠A = 65°Solution :
In Right angle △ABC,
we have,
=> ACB = 90°
=> CAB = 65°.
So,
=> ACB + CAB+ZCBA = 180° (By angle sum Property.)
=> 90° + 65° + CBA = 180°
=> 155° +CBA = 180°
=> CBA = 180° - 155°
=> CBA = 25°.
In △CDB,
=> CD is the altitude to AB.
So,
=> CDB = 90°
=> CBD = CBA = 25°.
So,
=> CBD + DCB = 180° (Angle sum Property.)
=> 90° +25° + DCB = 180°
=> 115° + DCB = 180°
=> DCB = 180° - 115°
=> DCB = 65°.
Now, in △ADC,
=> CD is the altitude to AB.
So,
=> ADC = 90°
=> CAD = CAB = 65°.
So,
=> ADC + CAD +DCA = 180° (Angle sum Property.)
=> 90° + 65° + DCA = 180°
=> 155° +DCA = 180°
=> DCA = 180° - 155°
=> DCA = 25°
Hence, we get,
DCA = 25°DCB = 65°CDB = 90°ACD = 25°ADC = 90°.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°.
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°
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°
I don't know how I can help you but I found this from another person. Hope this helps!
I don't know how I can help you but I found this from another person. Hope this helps!
Because f+e=c
Therefore ,
Step-by-step explanation:
Given A right triangle ABC as shown in figure where CD is an altitude of the triangle.
To prove that
Proofe: Given both are right triangle and both triangles have common angle B si same.
Therefore , two angles of two triangles are equal .
Hence, 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.Similarly , 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 .
The corresponding parts of two similar triangles are in equal proportion therefore , two proportion can be rewrite as
(I equation)
and (II equation)
Adding to both sides of firs equation
Because and ce can be substituted into the right side of equation wevcan write as
Applying the converse of distributive property we can write
Distributive property: a.(b+c)= a.c+a.b
Because f+e=
Hence proved.
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 .
It will provide an instant answer!