A triangular has 119 meters of frontage, and the other boundaries have lengths of 70 meters and 97 meters. What angles does the frontage make

Question

A triangular parcel of land has 119 meters of frontage, and the other boundaries have lengths of 70 meters and 97 meters. What angles does the frontage make with the two other boundaries? (Round your answers to one decimal place.)

Answer 1

The law of Cosines will provide the answer to this question.

 

That law states that in any triangle with sides A, B, and C , the Cos ∠A = (b² + c² - a²) ÷ 2bc

 

So, cos ∠A = (97² + 119² - 70²) / 2(97)(119) = 0.809

since the cosine is 0.809, we get the angle from a table which is 36^o.

 

We can then use the same law to calculate the other angles, since

cos ∠B = (a2 - b2 +c2) / 2ac

and ∠C = (a2 + b2 - c2) /2ab

which will result in B =54.6^o, and C = 89.4^o

Answer 2

Alexa,

To solve your problem, set up the problem using the Law of Cosines:

The component angles of the triangle, a, b, and c, are as follows:

Cos a = (b^2 + c^2 - a^2)/2(b)(c)

Cos a = (70^2 + 119^2 - 97^2) / 2 (70)(119)

Cos a = 9652 / 16660 = 0.58

a = cos^-1 (0.58) = 54.5 degrees

 

Cos b = (c^2 + a^2 - b^2)/ 2(a)(c)

Cos b = (14161 + 9409 - 4600)/ (2(119)(97)

Cos b = 0.8087

b = cos^-1 (0.8087) = 36 degree

 

Final angle, c, = 180 - 54.4 - 36 = 89.5, note: angles in a triangle must add up to 180.