Acute angle and triangle trig

Question

1) in triangle ABC <B=53 degrees, b=6.2 cm, and <C=47 degrees. Find the length of side a

 

2) A kayak leaves a dock on Lake Athatbasca, and heads due north for 2.8 km. At the same time, a second kayak travels in a direction N70 degrees E from the dock for 3.0km.

determine the distance between the kayaks, to the nearest tenth of a kilometre.

 

3)In triangle TUV, t=6.0 m, u=7.6 m, and v=8.0 m.  Solve the triangle. 

 

4) Two airplanes leave the Fort CHipewyan airport in Alberta at the same time. One airplane travels at 360 km/h. The other airplane travels at 430 km/h. About 30 minutes later they are 150 km apart. Determine the length between their paths.

Answer 1

For #3, you can use the law of cosines. If angle T is between sides u and v, then

 

t² = u² + v² - 2uv cos T

 

Similarly,

 

u² = t² + v² - 2tv cos U

v² = t² + u² - 2tu cos V

 

You know all 3 sides, so you can solve any two of these for two of the angles, then subtract from 180 to find the 3rd angle. To start with T:

 

(6.0m)² = (7.6m)² + (8.0m)² - 2(7.6m)(8.0m) cos T

36 = 57.76 + 64 - 121.6 cos T

-85.76 = -121.6 cos T

cos T = 0.70526

T = 45.15º

 

You should be able to find the other two angles from here!

 

Questions 2 and 4 are also applications of the law of cosines. If you draw pictures for each, labeling each side and opposite angle with the same letter (like t & T), then you can find your unknown with the law of cosines.

 

If you have further questions, please comment.

 

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4. I'm assuming you mistyped the problem, and it should be asking for the angle between the paths. The distance between them is given, 150km.

Both planes leave from the airport, lets call it A.

 

One plane travels at 360 km/h for 30 minutes = 0.5 hr for a total distance of 360*0.5=180 km. Label this endpoint B, and the side between A and B as c.

 

The other plane travels at 430 km/h for 30 minutes = 0.5 hr for a total distance of 430*0.5=215 km. Label this endpoint C, and the side between A and C as b.

 

The distance between B and C is side a, and is 150 km long, given in the problem.

 

Law of Cosines:

a² = b² + c² - 2bc cos A
150² = 215² + 180² - 2*215*180 cos A

 

You should be able to solve for A from here.

Answer 2

Hello, Amy

 

For # 1, we can use the law of sines.  If angle b= 53 and angle c=47, then, angle a= 80.  Using the law of sine

 

(sin 80)/a= (sin 53)/6.2       Using cross products

(6.2)(sin 80)= (a)(sin 53)     Finding sin, rounded to the thousandths.

(6.2)(.985)=(a)(.799)           Substituting the values 

       6.107= .799 a                Multiplying

            7.643= a.                  Dividing by .799

             7.6                           Rounded to the tenth place value.

 

For the second one, we can use the law of cosine, which is 

a^2=b^2+c^2-2bc cos a

a^2=(2.8)^2+(3)^2-(2)(2.8)(3) (cos 20)

a^2= 7.84+9- 16.8(cos 40)

a^2= 7.84+9-12.87

a^2= 3.97

a= 1.99  to the nearest tenth 2.0 km

 

Sorry, I can finish now, I will do it later. 

 

D. Y. Taylor