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第60頁

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解：如圖所示

$①在?Rt\triangle ABD?中，由勾股定理，得：$

$?BD=\sqrt {AB^2-AD^2}=3?$

$∴?CD=BC-BD=10?$

$在?Rt\triangle ADC?中，?AC=\sqrt {CD^2+AD^2}=2\sqrt {29}?$

$∴?\sin C=\frac {AD}{AC}=\frac {4}{2\sqrt {29}}=\frac {2\sqrt {29}}{29}?$

$? ②BD=\sqrt {AB2-AD2}=\sqrt {52-42}=3?$

$? CD=BC+BD=13+3=16?$

$? AC=\sqrt {AD2+CD2}=\sqrt {42+162}=4\sqrt {17}?$

$∴?sinC=\frac {AD}{AC}=\frac {\sqrt {17}}{17}?$

解：如圖所示

$∵?CD?是?Rt△ABC?斜邊上的中線$

$∴?CD=AD=BD=5，??AB=10，??AC=\sqrt {10^2-8^2}=6?$

$∴?∠ACD=∠A?$

$∴?sin∠ACD=sinA=\frac {BC}{AB}=\frac {4}{5}；??cos∠ACD=cosA=\frac {AC}{AB}=\frac {3}{5}?$

$?tan∠ACD=tanA=\frac {BC}{AC}=\frac {4}{3}?$

$?=\frac 12+\frac {\sqrt 2}2?$

$?=\frac {1+\sqrt 2}2?$

$?=(\frac {\sqrt 3}2)^2+(\frac 12)^2+1?$

$?=\frac 34+\frac 14+1?$

$?=2?$

$?=1-\frac 12?$

$?=\frac 12?$

$?=\frac 12+\frac {\sqrt 2}2-\frac {\sqrt 3}3?$

$解：?(1)?∵?tanα=\sqrt 3，?∴?α=60°.$

$??(2)sinα=\frac 12，??α=30°.$

$??(3)cosα=\frac {\sqrt 2}2，??α=45°$

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