Answer for a lot of points! In triangle △ABC, ∠C is a right angle and CD is the altitude to AB . Find the angles in △CBD and △CAD if m∠A = 65°
m∠DBC =
m∠DCB =
m∠CDB =
m∠ACD =
m∠ADC =

Given :

To find :

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,

see explanation

step-by-step explanation:

since the sequence is geometric there is a common ratio r between consecutive terms, that is

r = 4 ÷ - 2 = - 2

r = - 8 ÷ 4 = - 2

r = 16 ÷ 8 = - 2

the next term - 13 must be incorrect as it should be - 32 for the sequence to be geometric.

the first 4 terms are therefore geometric with common ratio r = - 2