Question

find the measures of the numbered angles in the kite.
m∠1 = 90°
m∠2 = 44°
m∠3 = □°

Explanation:

Step1: Analyze triangle ACD

Triangle ACD has right angle ∠1 = 90°, ∠2 = 44°. Sum of angles: 180°.

Step2: Calculate ∠3

∠3 = 180° - 90° - 44° = 46°? No, wait: The 46° is at vertex A, split by AC. So in triangle ACD, angle at A is 46°/2 = 23°? No, correction: Kite diagonals bisect vertex angles. ∠BAC = ∠DAC = 46°/2 = 23°. In right triangle ADC: ∠3 = 90° - ∠DAC = 90° - 23° = 67°? Wait no, original given m∠2=44°, so in right triangle ADC: ∠DAC + ∠2 + ∠1 = 180°, so ∠DAC = 180-90-44=46°? No, the top angle is 46°, so ∠BAD=46°, so ∠DAC=23°, then ∠3=90-23=67°? Wait no, the problem says m∠1=90°, m∠2=44°, so in triangle ADC: angles are 90°,44°, so the third angle (∠DAC) is 46°, which is half of ∠BAD=92°? Maybe the 46° is a typo, but using given m∠1=90, m∠2=44: in triangle ADC, ∠3 is the angle at A in triangle ABC? Wait no, the kite has AB=AD, CB=CD. Diagonals AC⊥BD (so ∠1=90). In triangle ABD, ∠BAD=46°, so ∠ABD=∠ADB=(180-46)/2=67°, but m∠2=44°, so in triangle CBD, CB=CD, so ∠CBD=∠CDB=44°, so ∠BCD=180-2*44=92°. Then in triangle ABC: right angle at 1, ∠ABC=∠ABD+∠CBD=67+44=111? No, wait the given m∠2=44°, so ∠2 is ∠CDB=44°, so in right triangle ADC: ∠DAC=90-44=46°, so ∠3=∠BAC=∠DAC=46°? No, the top angle is 46°, so ∠3 is half of (180-46)? No, the correct calculation using given m∠1=90 and m∠2=44: in triangle ADC, angles sum to 180: ∠3 + 90 + 44 = 180 → ∠3=46°?

Wait the problem's given values: m∠1=90°, m∠2=44°, so find m∠3. In the kite, AC bisects ∠BAD, and triangle ABC ≅ triangle ADC. In triangle ADC: ∠1=90°, ∠2=44°, so ∠DAC=180-90-44=46°. Since AC bisects ∠BAD=46°? No, ∠DAC=46°, so ∠3=∠DAC=46°?

Answer:

46