解直角三角形（双语课）[下学期]
图片预览
文档简介
Solving Right Triangle
Guangzhou Nan yang International School yongqian Gao
一、Teaching aims of the lesson:
1、To know the definition of solving right triangle
2、To understand the relationships among other five elements
3、To master how to solve right triangle
二、Teaching expressions:
1、To know how to solve right triangle
三、Difficult point in teaching:
1、To solve right triangles according to given conditions
四、Teaching procedures:
(Everybody, we have already learned four acute Trigonometric functions, today we are going to learn to solve right triangles ).
1、 Revision (Answer orally)
(1) As shown in Figure (1),
find out all the elements of the right triangle.
(∠A ,∠B , ∠C , a , b , c )
(2) In △ABC, ∠C＝90°, a= 5, and b=12, calculate the length of c and the values of four acute trigonometric functions of ∠A.
2、 Presentation
(1) The ways of solving math problem
①Given: a = b + c ,if a = 5 , b = 7 , c =_______;
②Given: , if a = 2, b = 2, ∠A=_______ ;
③Given: , if a = 2 , ∠A = 60°, c =_______;
(2) Besides right angle C , find out the equal relationships among other five elements .
(3) Think it over
Applying these relationships, if we know the other two elements. Can we find out three unknown elements
Example : if ∠A and c are given, then ∠B=________;
a =_______; b =_________; _________; _________.
(4) Training Exercises P21 1
(5) Example 1 In △ABC, ∠C is a right angle, the opposite
sides of ∠A, ∠B and ∠C are a, b and c, c=287.4, and
∠B=42°6′(42 degrees and 6 minutes).Solve this triangle.
Solution ① ∠A=90°－ 42°6′= 47°54′
② ∵ cos B =
∴ a = ccos B
=287.4×0.7420
≈213.3
③ ∵ sin B =
∴ b = csin B=287.4×0.6704≈192.7
(6) Training exercises P21 2 (1)
(answer: ∠A=9°36′, c=179.9 , b=177.4)
(7) Example 2 In Rt△ABC, a=104.0, and b=20.49. slove this triangle.
Solution ① ∵ tan A ==≈5.076,
∠A=78°51′（using calculator）
② ∠B=90°－78°51′=11°9′
③ ∵sin A=
∴ c ==≈106.0
(8) Training exercises P21 2 (2)
(answer: ∠A=69°13′, ∠B=20°47′, a=0.7786)
(9) Return to the question raised in the introduction part of this chapter.
(10) Summary:
本节课我们主要学习怎样解直角三角形.关键要抓住五个元素之间的等量关系,进行变形处理.并且要尽量采用原始数据, 最终结果按题目要求进行精确.(如果题目没有作要求,则边长取四位有效数字,角度精确到分)
(11) Homework: P32 (3)
C
B
A
注意：计算的最后结果，角度通常精确到分，边长通常保留四个有效数字。
Guangzhou Nan yang International School yongqian Gao
一、Teaching aims of the lesson:
1、To know the definition of solving right triangle
2、To understand the relationships among other five elements
3、To master how to solve right triangle
二、Teaching expressions:
1、To know how to solve right triangle
三、Difficult point in teaching:
1、To solve right triangles according to given conditions
四、Teaching procedures:
(Everybody, we have already learned four acute Trigonometric functions, today we are going to learn to solve right triangles ).
1、 Revision (Answer orally)
(1) As shown in Figure (1),
find out all the elements of the right triangle.
(∠A ,∠B , ∠C , a , b , c )
(2) In △ABC, ∠C＝90°, a= 5, and b=12, calculate the length of c and the values of four acute trigonometric functions of ∠A.
2、 Presentation
(1) The ways of solving math problem
①Given: a = b + c ,if a = 5 , b = 7 , c =_______;
②Given: , if a = 2, b = 2, ∠A=_______ ;
③Given: , if a = 2 , ∠A = 60°, c =_______;
(2) Besides right angle C , find out the equal relationships among other five elements .
(3) Think it over
Applying these relationships, if we know the other two elements. Can we find out three unknown elements
Example : if ∠A and c are given, then ∠B=________;
a =_______; b =_________; _________; _________.
(4) Training Exercises P21 1
(5) Example 1 In △ABC, ∠C is a right angle, the opposite
sides of ∠A, ∠B and ∠C are a, b and c, c=287.4, and
∠B=42°6′(42 degrees and 6 minutes).Solve this triangle.
Solution ① ∠A=90°－ 42°6′= 47°54′
② ∵ cos B =
∴ a = ccos B
=287.4×0.7420
≈213.3
③ ∵ sin B =
∴ b = csin B=287.4×0.6704≈192.7
(6) Training exercises P21 2 (1)
(answer: ∠A=9°36′, c=179.9 , b=177.4)
(7) Example 2 In Rt△ABC, a=104.0, and b=20.49. slove this triangle.
Solution ① ∵ tan A ==≈5.076,
∠A=78°51′（using calculator）
② ∠B=90°－78°51′=11°9′
③ ∵sin A=
∴ c ==≈106.0
(8) Training exercises P21 2 (2)
(answer: ∠A=69°13′, ∠B=20°47′, a=0.7786)
(9) Return to the question raised in the introduction part of this chapter.
(10) Summary:
本节课我们主要学习怎样解直角三角形.关键要抓住五个元素之间的等量关系,进行变形处理.并且要尽量采用原始数据, 最终结果按题目要求进行精确.(如果题目没有作要求,则边长取四位有效数字,角度精确到分)
(11) Homework: P32 (3)
C
B
A
注意：计算的最后结果，角度通常精确到分，边长通常保留四个有效数字。