In triangle δabc, ∠c is a right angle and cd is the height to ab. find the angles in ∠acd and ∠bcd if:

m∠a = 65°

Given:

m∠C = 90°, because ∠C is a right angle.

m∠D = 90°, because CD is the height to AB.

m∠A = α

Because the sum of angles in a triangle is 180°, therefore

m∠DBC + 90° + α = 180°

m∠DBC = 90° - α

Again, for the same reason,

m∠DCB + m∠DBC + 90° = 180°

m∠DCB + 90° - α + 90° = 180°

m∠DCB = α

For the same reason,

m∠ACD + 90° + α = 180°

m∠ACD = 90° - α

m∠ADC = 90° (by definition)

m∠CDB = 90° (by definition)

m∠DBC = 90° - α

m∠DCB = α

m∠CDB = 90°

m∠ACD = 90° - α

m∠ADC = 90°

Step-by-step explanation:

Learned this several times, also had this problem in my notes ;)

check above link.

graphically?   well, you graph them both, the solutions is where they both intersect each other, since there are two solutions, meaning they will meet twice, or namely the linear will run into the quadratic curve twice, and where they meet, those x,y coordinates are the solutions to the system.

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step-by-step explanation: