area of triangle w/ algebra

Question

how do you find the angle of a triangle with sides (6u-1) and (1u) and an area of 5cm^2 ?

 

 

 there was a question like this in a test i did but i couldn't work out the answer.. help?

Answer 1

The area of a triangle is base times height divided by 2.  The legs of the right triangle serve as the base and height. We need to first find the side lengths.  Then use law of sine to find the angles.

 

 

Set the equation for area of triangle.

 

u(6u - 1) / 2 = 5

 

 

Solve for u.

 

u(6u - 1) = 10

 

6u² - u = 10

 

 

Subtract 10 on both sides of the equation.

 

6u² - u - 10 = 0

 

 

Use the quadratic formula to solve for u.

 

u = (-b ± √(b² - 4ac)) / 2a

 

where:

a = 6

b = -1

c = -10

 

 

u = (1 ± √(1 + 240)) / 12

 

u = (1 ± √(241)) / 12

 

u = (1 ± 15.52) / 12

 

u = -1.21             and           u = 1.38

 

 

Accept the positive value of u:

 

u = 1.38

 

 

This makes the legs 1.38 cm and 7.28 cm.

 

Next, we use the Pythagorean theorem to find the hypotenuse, H.

 

H = √(1.38² + 7.28²)

H = 7.39 cm

 

 

Now we use law of sine.

 

smallest side / sin(smallest angle) = hypotenuse  / sin(90)

 

1.38 / sin(smallest angle) = 7.39

 

 

1.38 = 7.39 * sin(smallest angle)

 

0.1867 = sin(smallest angle)

 

 

Perform the inverse sine to find the angle.

 

10.76 = smallest angle

 

 

To find the other angle, subtract the smallest angle from 90 degrees.

 

90 - 10.76 = 79.24

Answer 2

A=½b·h

5=½u(6u-1)

3u²-1/2 u=5  Now we need to complete the square to then solve for u

3(u²-(1/6)u)=5

3(u²-(1/6)u+(1/44))=5+(1/48)

3(u-1/12)²=241/48

(u-1/12)²⁼241/144

(u-1/12)=±√(241)/12

u=±√(241)/12+1/12    I will only use u=(√(241)+1)/12 as the other value is negative.

 

6u-1=(√(241)-1)/12

arctan((6u-1)/u)= 79.26°

arctan(u/(6u-1))= 10.74°

 

I double checked my work by plugging in my u values into the formula for an area of a triangle and it still equaled 5 so this should be the answer.