# Junior High School Olympiad # Introduction to the Olympic Mathematical Competition or Mathematical Olympiad, referred to as Olympiad. Olympic Mathematics has a certain effect on teenagers' mental exercise, which can exercise their thinking and logic, and it plays a more profound role for students than ordinary mathematics. The following are the knowledge points of the Olympic Mathematics in the second day of junior high school: the right triangle is mainly explained, and everyone is welcome to read it.

First, solve the right triangle.

1. Definition: known edges and angles (two of which must have one side) → all unknown edges and angles.

2. Basis: ① Relationship between the two sides: review outline of junior high school mathematics

② Angle relation: A+B = 90.

③ Angular relation: the definition of trigonometric function.

Note: Try to avoid using intermediate data and division.

Second, the handling of practical problems.

1. Review outline of prone and elevation of junior high school mathematics

2. Azimuth and quadrant angle

3. Slope:

4. When both right triangles lack the conditions for solving right triangles, they can be solved by column equations.

Case analysis

It is known that port B is located 53.2 northeast of observation point A, and the distance BD from port B to observation point A in the north direction is 16km. A cargo ship sailed from port B to BC at a speed of 40km/h, and arrived at place C in 15 minutes. Now it is measured that point C is located at 79.8 northeast of observation point A. Find the length of the distance AC between the freighter and observation point A (accurate to 0. 1km). (Reference data: sin53.2 ≈0.80, cos53.2 ≈0.60, sin79.8 ≈0.98, cos79.8 ≈0. 18+08, tan26.6 ≈0.50.

Test center: the application of solving the right triangle-direction angle problem.

Analysis: according to sin∠DBA= in Rt△ADB, get the length of AB, then get the length of tan∠BAH=, then get the length of BH, then get the length of AH and CH, and then get the answer.

Solution: BC=40× = 10,

In Rt△ADB, sin∠DBA=, sin53.2 ≈0.8,

So AB= =20,

Point b is BH⊥AC, and the extension line of AC is H.

In Rt△AHB, ∠ bah = ∠ DAC-∠ dab = 63.6-37 = 26.6,

tan∠BAH=，0.5=，AH=2BH，

BH2+AH2=AB2, BH2+(2BH)2=202, BH=4, so AH=8,

In Rt△BCH, BH2+CH2=BC2, CH=2,

So AC=AH-CH=8 -2 =6 ≈ 13.4,

A: At this time, the distance AC between the freighter and observation point A is about 13.4km.

Comments: This topic mainly examines the direction angle in the right triangle, and finding the BH length according to the known right triangle structure is the key to solve the problem.