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五、常用三角计算和计算方法
1. 计算公式(见表1-12)
表1-12 计算公式
![](https://epubservercos.yuewen.com/25B71A/3590276104469801/epubprivate/OEBPS/Images/figure_0022_0038.jpg?sign=1739297384-qJCEY97iqCyvj249UqRfdzTSIwXmLWIm-0-b3a5fb873ddec620f9626979ac9e6063)
【例】 如图1-2所示,一个箱体两孔中心横向距离a=90mm,纵向距离b=70mm,求两孔的中心距离c是多少?
解:用公式
![](https://epubservercos.yuewen.com/25B71A/3590276104469801/epubprivate/OEBPS/Images/figure_0023_0040.jpg?sign=1739297384-OrP4MIss7arqCG8aODVqpB155Zxfj8v3-0-aafb7e64c1ba34e74d4f144ed2cdeb9f)
图1-2
查平方根表,则
。
所以两孔中心距离c=114mm。
2.30°、45°、60°角的三角函数值(表1-13)
表1-13 30°、45°、60°角的三角函数值
![](https://epubservercos.yuewen.com/25B71A/3590276104469801/epubprivate/OEBPS/Images/figure_0023_0043.jpg?sign=1739297384-zswcEEXaWqpMw1qrB4KRoTsCTOZ8k5Zd-0-126ebed6c81f9ef00e8b9462d8b2befd)
3. 三角函数表(见表1-14)
表1-14 三角函数表
![](https://epubservercos.yuewen.com/25B71A/3590276104469801/epubprivate/OEBPS/Images/figure_0024_0044.jpg?sign=1739297384-mnqYfKsEJwmWKinrLsGKubgo8vWlcC6D-0-7545d7710cd77ba437b4064592a96339)
![](https://epubservercos.yuewen.com/25B71A/3590276104469801/epubprivate/OEBPS/Images/figure_0025_0045.jpg?sign=1739297384-ysB1RwlWEPU9jQBxMlIeb0x0c63tANtq-0-64b29ca3f006c80092edb4e54069f330)
![](https://epubservercos.yuewen.com/25B71A/3590276104469801/epubprivate/OEBPS/Images/figure_0026_0046.jpg?sign=1739297384-PVYMagZvmVFEkbTLacZgYfxpd3SusrLY-0-fc03bc57b9033307800148311cea0a2d)
![](https://epubservercos.yuewen.com/25B71A/3590276104469801/epubprivate/OEBPS/Images/figure_0027_0047.jpg?sign=1739297384-OKEnKUVgP58pFq9SKJGvYxn9sBA0liL9-0-b00441a903bb6476a6839750da58d273)
![](https://epubservercos.yuewen.com/25B71A/3590276104469801/epubprivate/OEBPS/Images/figure_0028_0048.jpg?sign=1739297384-UHINQq9RIGEPI2mPPDOwMRlcIYSFjDDA-0-e60b9055106ba5f4486effd3890a1e0a)
![](https://epubservercos.yuewen.com/25B71A/3590276104469801/epubprivate/OEBPS/Images/figure_0029_0049.jpg?sign=1739297384-bdngT2hS3fXId2lT2DpJCj6ohLVr6uv9-0-6164cc9bf4a816c8417bd182add328cc)
![](https://epubservercos.yuewen.com/25B71A/3590276104469801/epubprivate/OEBPS/Images/figure_0030_0050.jpg?sign=1739297384-ih6TWbUzeiGawzM0E4uH0zWtoATdEXme-0-431dcf82de3c3bdafb478a8cd654c2d4)
![](https://epubservercos.yuewen.com/25B71A/3590276104469801/epubprivate/OEBPS/Images/figure_0031_0051.jpg?sign=1739297384-IaXjuIcfgi2FQOlOMmvsd5BgZjUGRSRd-0-302dfffaceda91d1176556aceee4d439)
![](https://epubservercos.yuewen.com/25B71A/3590276104469801/epubprivate/OEBPS/Images/figure_0032_0052.jpg?sign=1739297384-wP0EZf032tdri2YAQxZWGcBomOdnUopa-0-9bf1551f221174666cff92fa97ce1dac)
![](https://epubservercos.yuewen.com/25B71A/3590276104469801/epubprivate/OEBPS/Images/figure_0033_0053.jpg?sign=1739297384-hQIJXmQho5kEzfpUc70fmZGqz9G6YXWC-0-bdbc6773fd52403c5cb43ceb96dcaffe)
![](https://epubservercos.yuewen.com/25B71A/3590276104469801/epubprivate/OEBPS/Images/figure_0034_0054.jpg?sign=1739297384-73sL2hq7rN3NTOAOgQcZJ3vpSDNmS9pv-0-7ee4fc1d6bf0bedd51e7f16db2eaf6ee)
![](https://epubservercos.yuewen.com/25B71A/3590276104469801/epubprivate/OEBPS/Images/figure_0035_0055.jpg?sign=1739297384-sah0mgfvb0QFfvqBwAs17IYg4NAa26p4-0-5d4dee98180299e85fb9b69958252534)
![](https://epubservercos.yuewen.com/25B71A/3590276104469801/epubprivate/OEBPS/Images/figure_0036_0056.jpg?sign=1739297384-egexHGd5m1ThvUhPHnEptPozFta3G1ym-0-0e3950a55573bb92b27076ea2572fba0)
![](https://epubservercos.yuewen.com/25B71A/3590276104469801/epubprivate/OEBPS/Images/figure_0037_0057.jpg?sign=1739297384-e9UuRNaTVHQbjZHObahSiMJh31ETQmNe-0-569dcdac14f3cd6833ccdd9f2d43e74c)
![](https://epubservercos.yuewen.com/25B71A/3590276104469801/epubprivate/OEBPS/Images/figure_0038_0058.jpg?sign=1739297384-XalzcOette5yUOIjrlvWZ4b7q1zzvoTf-0-d1cdc8c29cb6e2f01e9da2bb54f6c357)
![](https://epubservercos.yuewen.com/25B71A/3590276104469801/epubprivate/OEBPS/Images/figure_0039_0059.jpg?sign=1739297384-Eh3Pd0zibl7vqZ8AjNrj4uGLgJEcYLRJ-0-4d5dc192b8be630aacf6a02c13bfa31c)
![](https://epubservercos.yuewen.com/25B71A/3590276104469801/epubprivate/OEBPS/Images/figure_0040_0060.jpg?sign=1739297384-DFbLvUydXmbdOVfXqETS88PpcjORxndU-0-55f879c0e11072323c00e22ca5e6439b)
![](https://epubservercos.yuewen.com/25B71A/3590276104469801/epubprivate/OEBPS/Images/figure_0041_0061.jpg?sign=1739297384-SF6BtAndEivBiRbv2EECOGZLDXEMQ5Xg-0-5282c6fd6f05b4e42f177ec2d2a49a5c)
![](https://epubservercos.yuewen.com/25B71A/3590276104469801/epubprivate/OEBPS/Images/figure_0042_0062.jpg?sign=1739297384-4mGAJmaFm1zshPLrmuArrqCRajVJimHH-0-f06c1d391f8b22090a85ce1884c693c9)
![](https://epubservercos.yuewen.com/25B71A/3590276104469801/epubprivate/OEBPS/Images/figure_0043_0063.jpg?sign=1739297384-svmauYNhoYXGHwbTMwTLaCy0x92mpTA2-0-70e828bc25ea2434b7af74c0d6f7ac15)
![](https://epubservercos.yuewen.com/25B71A/3590276104469801/epubprivate/OEBPS/Images/figure_0044_0064.jpg?sign=1739297384-2DPqI5JrkqVCeAuiLWyszEMOMVfvdIai-0-caba0256a6a84283f1c3399f78d36ec5)
![](https://epubservercos.yuewen.com/25B71A/3590276104469801/epubprivate/OEBPS/Images/figure_0045_0065.jpg?sign=1739297384-miBtBus2LgL4BpIKW1dOV15OgTxKw6FH-0-82d38e14cf077a220e5b6d8ae8324986)
![](https://epubservercos.yuewen.com/25B71A/3590276104469801/epubprivate/OEBPS/Images/figure_0046_0066.jpg?sign=1739297384-GoEq4bkOjmdGEUpR1uBw1wPcG6PQrjOE-0-769ea424b78a74e6061f4c0b44c9d48b)
用法说明:
本表的角度间隔4′,若遇到4′的中间数(如34°35′40″)的角度,可以用比例法进行修正。
【例】 求30°15′的正弦值(sin35°15′)。
解:先查出sin30°12′=0.50302
sin30°16′=0.50403
取sin30°12′与sin30°16′的差值除以4,可以作为1′的正弦值。
![](https://epubservercos.yuewen.com/25B71A/3590276104469801/epubprivate/OEBPS/Images/figure_0046_0067.jpg?sign=1739297384-yxlzrO37jonsFtTmNP2IwbUsAEUn0FUJ-0-6162c86eba090f0e7dc650b6803c0ac6)
sin30°15′=sin30°16′-sin1′=0.50403-0.00023=0.50380
【例】 求24°35′40″的正切值(tan24°35′40″)。
解:先查出tan24°32′=0.45643
tan24°36′=0.45784
则,说明当角度增加1′(60″)时,正切值约增加0.00035,如角度是20′,假设20″的正切值为
x,这时可以列出比例式:
![](https://epubservercos.yuewen.com/25B71A/3590276104469801/epubprivate/OEBPS/Images/figure_0047_0069.jpg?sign=1739297384-NMKmjfcUVFdTXQKAfXzAR01aaXoF1KiU-0-21a5f5d8099809f2182237ccd797a0bd)
x≈0.00012
所以tan24°35′40″=tan24°36′-tan20″=0.45784-0.00012=0.45772。
【例】 已知某角的正切值等于0.5824,求该角度大小。
解:从表上“正切tan”一栏查出与0.5824相近的函数值0.58201和0.58357
0.58357-0.58201=0.00156
0.58201对应的角度是30° 12′,058357 对应的角度是30° 16′,说明当正切值增加0.00156时,角度增加4′。现在某角的正切值为0.5824,比30° 12′的正切值0.58201 增加0.00039(0.5824-0.58201=0.00039),可以根据比例式求出角度的增加值x:
![](https://epubservercos.yuewen.com/25B71A/3590276104469801/epubprivate/OEBPS/Images/figure_0047_0070.jpg?sign=1739297384-TAlVgXC53AO8oSh2AXc9MeByHFMcgzim-0-fb3908039270c5694398b15761c9498f)
所以正切值为0.5824的角度为30°12′+1′=30°13′。