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° m∠dbc = ? m∠dcb = ? m∠cdb = ? m∠acd = ? m∠adc = ?

Part 1) m∠DBC=25°

Part 2) m∠DCB=65°

Part 3) m∠CDB=90°

Part 4) m∠ACD=25°

Part 5) m∠ADC=90°

Step-by-step explanation:

see the attached figure to better understand the problem

step 1

Find the measure of angle DBC

we know that

The sum of the interior angles of a triangle must be equal to 180 degrees

In the right triangle ABC

m∠A+m∠B+m∠C=180° ----> equation A

we have

m∠A=65° ----> given problem

m∠C=90° ----> given problem

Substitute the given values in the equation A and solve for m∠B

65°+m∠B+90°=180°

m∠B+155°=180°

m∠B=180°-155°

m∠B=25°

Remember that the measure of Angle B is equal to say the measure of angle DBC

so

m∠B=m∠DBC

therefore

m∠DBC=25°

step 2

Find the measure of angle DCB and angle CDB

In the right triangle DBC

The sum of the interior angles of a triangle must be equal to 180 degrees

m∠DBC+m∠DCB+m∠CDB=180°

we have

m∠DBC=25°

m∠CDB=90° ----> is a right angle (CD is the height to AB)

substitute the values and solve for m∠DCB

25°+m∠DCB+90°=180°

m∠DCB+115°=180°

m∠DCB=180°-115°=65°

step 3

Find the measure of angle ACD

we know that

m∠ACD+m∠DCB=90° -----> by complementary angles

we have

m∠DCB=65°

substitute the value

m∠ACD+65°=90°

m∠ACD=90°-65°=25°

step 4

Find the measure of angle ADC

m∠ADC=90° ----> is a right angle (CD is the height to AB)

the first one is x^2+2x+2

the second one is x^2+4

the last one is x^2-4x+8

i dont understand

step-by-step explanation: