题目内容
如图,BD平分∠ABC,CD平分角ACB.
(1)若∠A=50°,求∠BDC的度数;
(2)若∠A=α,试用α的式子表示∠BDC.
(1)若∠A=50°,求∠BDC的度数;
(2)若∠A=α,试用α的式子表示∠BDC.
(1)∵BD平分∠ABC,CD平分∠ACB,
∴∠DBC=
∠ABC,∠DCB=
∠ACB,
∴∠BDC=180°-∠DBC-∠DCB=180°-
(∠ABC+∠ACB),
∵∠ABC+∠ACB=180°-∠A,
∴∠BDC=180°-
(180°-∠A)
=90°+
∠A,
=90°+
×50°
=115°;
(2)∵∠BDC=90°+
∠A,
∴∠BDC=90°+
α.
∴∠DBC=
1 |
2 |
1 |
2 |
∴∠BDC=180°-∠DBC-∠DCB=180°-
1 |
2 |
∵∠ABC+∠ACB=180°-∠A,
∴∠BDC=180°-
1 |
2 |
=90°+
1 |
2 |
=90°+
1 |
2 |
=115°;
(2)∵∠BDC=90°+
1 |
2 |
∴∠BDC=90°+
1 |
2 |
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