2026年补充习题江苏七年级数学上册苏科版第99页答案
5. 计算:
(1) $45°18' - 36°56'$;
(2) $3 × 15°30'$。

答案

解:
(1) 原式$=44°78' - 36°56'$
$=8°22'$
(2) 原式$=45°90'$
$=46°30'$
6. 根据图形填空:
(1)$∠ AOB = ∠ AOD + \_\_\_\_\_\_ = ∠ \_\_\_\_\_\_ + ∠ BOC$;
(2)$∠ BOD - ∠ BOC = \_\_\_\_\_\_$;
(3)$∠ AOD + ∠ BOC = \_\_\_\_\_\_ - \_\_\_\_\_\_$;

答案

解:
(1)$∠ AOB = ∠ AOD + \boldsymbol{∠ DOB} = ∠ \boldsymbol{AOC} + ∠ BOC$;
(2)$∠ BOD - ∠ BOC = \boldsymbol{∠ COD}$;
(3)$∠ AOD + ∠ BOC = \boldsymbol{∠ AOB} - \boldsymbol{∠ COD}$。
7. 如图,$∠ BAD=∠ CAE$.
(1)写出图中另一对相等的角,并说明理由.
(2)若$∠ BAC=106°,∠ CAE=24°48'$,求$∠ CAD$的大小.

答案

解:
(1) $∠ CAD = ∠ BAE$,理由如下:
$\because ∠ BAD = ∠ CAE$,
$\therefore ∠ BAD + ∠ EAD = ∠ CAE + ∠ EAD$,
即 $∠ BAE = ∠ CAD$。
(2) $\because ∠ BAC = 106°$,$∠ CAE = 24° 48'$,
$\therefore ∠ BAE = ∠ BAC - ∠ CAE = 106° - 24° 48' = 81° 12'$,
又$\because ∠ CAD = ∠ BAE$,
$\therefore ∠ CAD = 81° 12'$。