Mathematics, 10.07.2019 20:30 bskyeb14579

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°

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Answer from: turtlesage21

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°.

Answer from: michaelswagout

In triangle CBD

$\angle CBD=25^{\circ}, \angle CDB=90^{\circ}, \angle BCD=65^{\circ}$

In triangle CAD

$\angle CAD=65^{\circ}, \angle CDA=90^{\circ}, \angle ACD=25^{\circ}$

Step-by-step 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 from: Quest

56 ur welcome:

Answer from: Quest

i believe your answer is 99.7% of 7-year-old children are between 46.6 inches and 49.5 inches tall.

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