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解直角三角形

解直角三角形. 6.5 应用举例 ( 二 ). B. 解直角三角形依据. c. a. (1) 三边之间的关系:.   a 2 + b 2 = c 2 ( 勾股定理 ). C. A. (2) 锐角之间的关系:. b. ∠A + ∠ B = 90°. (3) 边角之间的关系:. cosA =. 其中 A 可换 成 B. sinA =. cotA=. tan=. 三个. 这三个关系式中,每个关系式都包含 元素,知其中 元素就可以求出. 两个. 第三个元素.

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解直角三角形

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  1. 解直角三角形 6.5 应用举例(二)

  2. B 解直角三角形依据 c a (1)三边之间的关系:   a2+b2=c2(勾股定理) C A (2)锐角之间的关系: b ∠A+∠B=90° (3)边角之间的关系: cosA = 其中 A可换 成B sinA= cotA= tan= 三个 这三个关系式中,每个关系式都包含元素,知其中 元素就可以求出 两个 第三个元素

  3. 例1 如图,厂房屋顶人字架(等腰三角形)的跨度为10米,∠A=26º,求中柱BC(C为底边中点)和上弦AB的长(精确到0.01米). (tan26º=0.4877,cos26º=0.8988) ? ? 26º 5

  4. 例1 如图,厂房屋顶人字架(等腰三角形)的跨度为10米,∠A=26º,求中柱BC(C为底边中点)和上弦AB的长(精确到0.01米). (tan26º=0.4877,cos26º=0.8988) 解:∵tgA=BC/AC, ∴BC=ACtgA =5×tg26º 26º ≈2.44(米). 5 ∵cosA=AC/AB, ∴AB=AC/cosA =5/cos26º =5/0.8989 ≈5.56(米). 答:中柱BC约长2.44米,上弦AB约长5.56米.

  5. =1.2799 tg52º 例2 为测量松树AB的高度,一个人站在距松树15米的E处,测得仰角∠ACD=52º,已知人的高度是1.72米,求树高(精确到0.01米). ? AB=AD+BD 52º 1.72 ? 15 1.72

  6. 例2 为测量松树AB的高度,一个人站在距松树15米的E处,测得仰角∠ACD=52º,已知人的高度是1.72米,求树高(精确到0.01米). =1.2799 tg52º 解:在Rt⊿ACD中, tgC=AD/CD, ∴AD=CDtgC=BEtgC =15×tg52º =15×1.2799≈19.20(米). ∴AB=AD+BD=19.20+1.72 =20.92(米). 答:树高20.92米.

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