Triangle with base line of 6 metres, connecting to 2 equal sides with angles of 10 degrees
at the connections. What are the lengths of the remaining 2 sides?
To find the hypotenuse of a right triangle, we use the formula c^2 = a^2 + b^2.
That is derived from the general forumula c^2 = a^2 + b^2 - (2 * a * b * cos C).
It works because the cosine of a 90 degree angle is 0, so the last term is 0, due to multiplying by 0.
To solve the problem before us, we need that general formula.
We know c, the would-be hypotenuse; it is the 6m base.
The other two sides, a and b, are equal length, so we will refer to both lengths as a.
Angle C is the angle opposite the base. We are given that there are two 10 degree angles, so angle C must be 160 degrees.
c^2 = a^2 + b^2 - (2 * a * b * cos C)
6^2 = a^2 + a^2 - (2 * a * a * cos 160) Remember, a is the length of each of the other two sides.
36 = (2 * a^2) - (2 * a^2 * cos 160) We factor out (2 * a^2)
36 = (2 * a^2) * (1 - cos 160) Next, divide by (1 - cos 160)
36 / (1 - cos 160) = 2 * a^2 Now, divide by 2
18 / (1 - cos 160) = a^2 The final step to simplify is to take the square root
sqrt(18 / (1 - cos 160)) = a Perform the calculations and you have the length of the other two sides.
a = sqrt (18 / (1 - (-0.9397))) = sqrt (18 / 1.9397)
a = sqrt (9.2798) = 3.046m
Adding the lengths of the two sides gives 6.093, barely longer than the base, but that shows the triangle is extremely short.
That's why the two angles at either end of the base are only 10 degrees.