Question

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

Answer 1

Given :-

• ∠C = 90°
• CD is the altitude to AB.
• ∠A = 65° .

Solution :-

in Right angle ∆ABC , we have ,

→ ∠ACB = 90°

→ ∠CAB = 65° .

So,

→ ∠ACB + ∠CAB + ∠CBA = 180° (By angle sum Property.)

→ 90° + 65° + ∠CBA = 180°

→ 155° + ∠CBA = 180°

→ ∠CBA = 180° - 155°

→ ∠CBA = 25° .

Now, in ∆CDB ,

→ CD is the altitude to AB.

So,

→ ∠CDB = 90°

→ ∠CBD = ∠CBA = 25° .

So,

→ ∠CDB + ∠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, from all above (Also image.) we get,

• ∠DBC = 25°
• ∠DCB = 65°
• ∠CDB = 90°
• ∠ACD = 25°
• ∠ADC = 90° .

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