Answer:
Step-by-step explanation:
1). m∠AEC = m∠AEB + m∠BEC
= 21° + 37°
= 58°
2). m∠BED = m∠BEC + m∠CED
= 37° + 44°
= 81°
3). m∠IKF = m∠IKH + m∠HKG + m∠GKF
= m∠IKH + m∠HKG + m∠IKH [Since, ∠IKH ≅ ∠GKF]
= 2∠IKH + m∠HKG
103° = 2∠IKH + 41°
2(∠IKH) = 103 - 41
m(∠IKH) = 31°
4). m∠AED = m∠AEB + m∠BEC + m∠CED
= 21° + 37° + 44°
= 102°
5). m∠JKG = 108°
m∠JKG = m∠JKI + m∠IKH + m∠HKG
108° = m∠JKI + 31° + 41°
m∠JKI = 108° - 72°
m∠JKI = 36°
6). m∠HKF = m∠GKF + m∠HKG
= m∠IKH + m∠HKG [Since, m∠GKF = m∠IKH]
= 31° + 41°
= 72°
7). m∠NQO = m∠MQN = 64°
8). m∠JKF = m∠JKI + m∠IKF
= 36° + 103°
= 139°
8). m∠MQO = 2(m∠NQO)
= 2(64)°
= 128°
9). m∠LQO = 156°
m∠LQM = m∠LQO - m∠MQO
= 156° - 128°
= 28°
10. m∠NQP = m∠NQO + m∠OQP
= 64° + m∠LQM [Since ∠OQP ≅ ∠LQM]
= 64° + 28°
= 92°
Answer:
i know the answer
Step-by-step explanation:
68% because it’s between them
Answer:
C. Triangle BAC is congruent to triangle FDE by AAS
Step-by-step explanation:
BAC names the vertices in the order longest-side, shortest-side. That same order is FDE in the other triangle, eliminating choiced B and D. The triangles are not right triangles, eliminating choice A.
The only viable answer choice is C.
No specific sides are shown as being congruent, but two angles are, so we could claim congruence by ASA or AAS. Answer choice C uses the latter.
Answer:
girll how is this college math im in 8th grade and n learned this in 1st grade loll
Step-by-step explanation:
its 15 m I THINK DONT TAKE MY WORD ON IT but tytyty for the points
