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°

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°.

r = 78539458390583095803458309458039485093485 or just 3

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