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6. 在$\triangle ABC$中,$AD为边BC$上的高,$∠ABC = 30^{\circ}$,$∠CAD = 20^{\circ}$,则$∠BAC$是
80 或 40
度。答案
6.80 或 40
7. 如图 13 - 14,$∠A = 70^{\circ}$,$∠B = 26^{\circ}$,$∠C = 20^{\circ}$,则$∠BDC$的度数为
$116^{\circ}$
。答案
7.$116^{\circ}$ 提示:延长 BD 交 AC 于点 H,
则$∠AHB = 180^{\circ} - (∠A + ∠B) = 180^{\circ} - 96^{\circ} = 84^{\circ}$.
又∵$∠C = 20^{\circ}$,
∴$∠CDH = 84^{\circ} - 20^{\circ} = 64^{\circ}$,
∴$∠BDC = 180^{\circ} - ∠CDH = 116^{\circ}$.
则$∠AHB = 180^{\circ} - (∠A + ∠B) = 180^{\circ} - 96^{\circ} = 84^{\circ}$.
又∵$∠C = 20^{\circ}$,
∴$∠CDH = 84^{\circ} - 20^{\circ} = 64^{\circ}$,
∴$∠BDC = 180^{\circ} - ∠CDH = 116^{\circ}$.
8. 如图 13 - 15,在$\triangle ABC$中,$BE⊥AC于点E$,$AF是∠CAB$的平分线,交$BE于点F$,$∠C = 78^{\circ}$,$∠CBA = 38^{\circ}$,求$∠AFB$的度数。
解：∵$∠C = 78^{\circ}$,$∠CBA = 38^{\circ}$,
∴$∠CAB = 180^{\circ} - ∠C - ∠CBA = 180^{\circ} - 78^{\circ} - 38^{\circ} = 64^{\circ}$.
∵ AF 是$∠CAB$的平分线,
∴$∠EAF = ∠FAB = \frac{1}{2}∠CAB = 32^{\circ}$.
∵$BE \perp AC$,∴$∠AEB = 90^{\circ}$.
∴$∠AFB = ∠EAF + ∠AEB = 32^{\circ} + 90^{\circ} =
解：∵$∠C = 78^{\circ}$,$∠CBA = 38^{\circ}$,
∴$∠CAB = 180^{\circ} - ∠C - ∠CBA = 180^{\circ} - 78^{\circ} - 38^{\circ} = 64^{\circ}$.
∵ AF 是$∠CAB$的平分线,
∴$∠EAF = ∠FAB = \frac{1}{2}∠CAB = 32^{\circ}$.
∵$BE \perp AC$,∴$∠AEB = 90^{\circ}$.
∴$∠AFB = ∠EAF + ∠AEB = 32^{\circ} + 90^{\circ} =
122^{\circ}
$.答案
8.∵$∠C = 78^{\circ}$,$∠CBA = 38^{\circ}$,
∴$∠CAB = 180^{\circ} - ∠C - ∠CBA = 180^{\circ} - 78^{\circ} - 38^{\circ} = 64^{\circ}$.
∵ AF 是$∠CAB$的平分线,
∴$∠EAF = ∠FAB = \frac{1}{2}∠CAB = 32^{\circ}$.
∵$BE \perp AC$,∴$∠AEB = 90^{\circ}$.
∴$∠AFB = ∠EAF + ∠AEB = 32^{\circ} + 90^{\circ} = 122^{\circ}$.
∴$∠CAB = 180^{\circ} - ∠C - ∠CBA = 180^{\circ} - 78^{\circ} - 38^{\circ} = 64^{\circ}$.
∵ AF 是$∠CAB$的平分线,
∴$∠EAF = ∠FAB = \frac{1}{2}∠CAB = 32^{\circ}$.
∵$BE \perp AC$,∴$∠AEB = 90^{\circ}$.
∴$∠AFB = ∠EAF + ∠AEB = 32^{\circ} + 90^{\circ} = 122^{\circ}$.
9. 如图 13 - 16,已知点$O是\triangle ABC$的两条角平分线的交点。
(1)若$∠A = 30^{\circ}$,则$∠BOC$的度数是______
(2)若$∠A = 60^{\circ}$,则$∠BOC$的度数是______
(3)若$∠A = n^{\circ}$,则$∠BOC$的度数是多少？试用学过的知识说明理由。
(1)若$∠A = 30^{\circ}$,则$∠BOC$的度数是______
$105^{\circ}$
。(2)若$∠A = 60^{\circ}$,则$∠BOC$的度数是______
$120^{\circ}$
。(3)若$∠A = n^{\circ}$,则$∠BOC$的度数是多少？试用学过的知识说明理由。
$∠BOC = \frac{1}{2}n^{\circ} + 90^{\circ}$.
理由如下:在$△ABC$中,$∠A + ∠ABC + ∠ACB = 180^{\circ}$. 在$△BOC$中,$∠BOC + ∠OBC + ∠OCB = 180^{\circ}$.
∵ BO,CO 分别是$∠ABC$和$∠ACB$的平分线,
∴$∠ABC = 2∠OBC$,$∠ACB = 2∠OCB$.
∴$∠BOC + \frac{1}{2}∠ABC + \frac{1}{2}∠ACB = 180^{\circ}$.
又∵ 在$△ABC$中,$∠A + ∠ABC + ∠ACB = 180^{\circ}$,
∴$\frac{1}{2}∠ABC + \frac{1}{2}∠ACB = 90^{\circ} - \frac{1}{2}∠A$.
∴$∠BOC = \frac{1}{2}∠A + 90^{\circ}$.
∴ 若$∠A = n^{\circ}$,则$∠BOC = \frac{1}{2}n^{\circ} + 90^{\circ}$.
理由如下:在$△ABC$中,$∠A + ∠ABC + ∠ACB = 180^{\circ}$. 在$△BOC$中,$∠BOC + ∠OBC + ∠OCB = 180^{\circ}$.
∵ BO,CO 分别是$∠ABC$和$∠ACB$的平分线,
∴$∠ABC = 2∠OBC$,$∠ACB = 2∠OCB$.
∴$∠BOC + \frac{1}{2}∠ABC + \frac{1}{2}∠ACB = 180^{\circ}$.
又∵ 在$△ABC$中,$∠A + ∠ABC + ∠ACB = 180^{\circ}$,
∴$\frac{1}{2}∠ABC + \frac{1}{2}∠ACB = 90^{\circ} - \frac{1}{2}∠A$.
∴$∠BOC = \frac{1}{2}∠A + 90^{\circ}$.
∴ 若$∠A = n^{\circ}$,则$∠BOC = \frac{1}{2}n^{\circ} + 90^{\circ}$.
答案
9.(1)$105^{\circ}$ (2)$120^{\circ}$
(3)$∠BOC = \frac{1}{2}n^{\circ} + 90^{\circ}$.
理由如下:在$△ABC$中,$∠A + ∠ABC + ∠ACB = 180^{\circ}$. 在$△BOC$中,$∠BOC + ∠OBC + ∠OCB = 180^{\circ}$.
∵ BO,CO 分别是$∠ABC$和$∠ACB$的平分线,
∴$∠ABC = 2∠OBC$,$∠ACB = 2∠OCB$.
∴$∠BOC + \frac{1}{2}∠ABC + \frac{1}{2}∠ACB = 180^{\circ}$.
又∵ 在$△ABC$中,$∠A + ∠ABC + ∠ACB = 180^{\circ}$,
∴$\frac{1}{2}∠ABC + \frac{1}{2}∠ACB = 90^{\circ} - \frac{1}{2}∠A$.
∴$∠BOC = \frac{1}{2}∠A + 90^{\circ}$.
∴ 若$∠A = n^{\circ}$,则$∠BOC = \frac{1}{2}n^{\circ} + 90^{\circ}$.
(3)$∠BOC = \frac{1}{2}n^{\circ} + 90^{\circ}$.
理由如下:在$△ABC$中,$∠A + ∠ABC + ∠ACB = 180^{\circ}$. 在$△BOC$中,$∠BOC + ∠OBC + ∠OCB = 180^{\circ}$.
∵ BO,CO 分别是$∠ABC$和$∠ACB$的平分线,
∴$∠ABC = 2∠OBC$,$∠ACB = 2∠OCB$.
∴$∠BOC + \frac{1}{2}∠ABC + \frac{1}{2}∠ACB = 180^{\circ}$.
又∵ 在$△ABC$中,$∠A + ∠ABC + ∠ACB = 180^{\circ}$,
∴$\frac{1}{2}∠ABC + \frac{1}{2}∠ACB = 90^{\circ} - \frac{1}{2}∠A$.
∴$∠BOC = \frac{1}{2}∠A + 90^{\circ}$.
∴ 若$∠A = n^{\circ}$,则$∠BOC = \frac{1}{2}n^{\circ} + 90^{\circ}$.
