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Math problems of trigonometric function in grade three
Note: your drawing is not very accurate. 30 degrees west of north means 30 degrees west of north, and 60 degrees east of north means 60 degrees east of north.

Known AB=40km, AC=8√3km, ∠ BAC = 30+60 = 90, ∠ CAM = 90-60 = 30.

∴ BC=√(AB? +AC? )= 16√7

In the right triangle ABC:

cos∠ACB = AC/BC = 8√3/( 16√7)=√2 1/ 14

sin∠ACB = AB/BC = 40/( 16√7)= 5√7/ 14

Extend BC to Ann to D.

∠∠ACB =∠D+∠DAC

∴sin(∠d)= sin(∠ACB-∠DAC)= sin∠ACB * cos∠DAC-con∠ACB sin∠DAC

=5√7/ 14 * √3/2 - √2 1/ 14 * 1/2

=√2 1/7

sin∠ACD = sin( 180-∠ACB)= sin∠ACB

According to sine theorem

AD/sin∠ACD=AC/sin∠D

AD = 8√3/(√2 1/7)* 5√7/ 14 = 20k m

∫AM = 19.5,AN= 19.5+ 1=20.5

Am & ltAD & ltart 1 (before vowels and phonemes)

This ship can just reach the pier MN and the pier.