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Trigonometry help!!!!!!!!!!!!!!!

+2 votes
In the triangle ABC, if side a is 3, side b is 4 and side c is 5, what is the cosine of the smallest angle?
asked Feb 16, 2013 in TRIGONOMETRY by andrew Scholar

1 Answer

+3 votes

Given that side a is 3, side b is 4 and side c is 53

Find the largest angle of the triangle first. This will be / C because the longest side is C.

Law of cosines: c2 = a2 + b2 -2abcosC

52 = 32 + 42 - 2(3)(4)cosC.

25 = 9 + 16 - 24cosC

25 = 25 - 24cosC

Subtract 25 from each side.

-24cosC = 0

Divide each side by -24.

cosC = 0

Trigonometric table in cos90° = 0

cosC = cos90 then /C = 90°

Now that we have an angle, we can switch to the law of sines.

Find / B: sinB/b = sinC/c

sinB/4 = sin90/5

sinB/4 = 1/5

Multiply each side by 4.

sinB = 4/5 = 0.8

Trigonometric table in sin53° = 0.8

sinB = sin53° then /B = 53°

Total of the three angles in the triangle is 180°.

So, A+B+C = 180°

Substitute C = 90° and B = 53° in the equation.

A + 53° + 90° = 180°

A + 143° = 180°

Subtract 143° from each side.

/A = 37°

Therefore /A = 37°, /B = 53° and /C = 90°.

answered Feb 16, 2013 by britally Apprentice

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