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The eighth grade mathematics chapter 65438 +04 solves problems.
97. The sunshine is warm. Not much to say.

98. (Symbol representing angle. Because I can't play corner yards)

Mark the angle ABE as angle 1 on the diagram. The EBD angle is 32. The DBC angle is 3 degrees.

The angle BEC is 4 degrees and the angle BDA is 5 degrees.

Because AB=AD, 5 = 1+ 2.

υ5 =υ3+υc (triangle outer angle = sum of two inner angles)

That is, 5 = 3+(90-a), then1+2 = 3+90-a.

a =?3-? 1-?2+90

Similarly, i4 = ia+I1,CB=CE, so i4 = I2+i3.

Deco A+ CuO 1 = CuO 2+ CuO 3, then CuO A= CuO 2+ CuO 3- CuO 1.

You have 2+3- 1 = 3- 1-2+90.

Then 2* 32 = 90

Well, 2=45. That is, the angle EBD is 45 degrees.

99。 ( 1)

Just south of island B is point D. AD vertical BD

Angle BCD = 60. Then BC=2*CD=40, then CD = 20 (right angle side facing 30 degrees in RT triangle = half of hypotenuse).

Well, ABD=90-30=60. CBD=30。 Then ABC=60-30=30

A=30。 AC=BC=40 nautical miles. Time T= distance/speed =40/ 10=4 hours (11:30+4 =15: 30) The time to arrive at point C is 3: 30 pm.

(2)

T=20/ 10=2 (hours), so it's 5: 30 pm when we get to D.

100。

Connect AF. Then AF=CF (middle vertical line property) because A= 120 AB=AC. Then B= C=30.

BAF= 120-30=90。

So ... AF=( 1/2)BF. And AF=CF, so CF=( 1/2)BF is BF=2CF.

I hope you understand the above questions.