What is the base length of an isosceles triangle with two legs of 5 inches the obtuse angle of 120 degrees and the two base angles of 30 degrees?

Answer 1

First, draw the triangle ABC. Denote with A and B the two base angles, and with C the obtuse angle. Denote with a the side opposite the angle A, and with b the opposite of the angle B (since these two sides are equal, you can denote both of them with a), and with c the side opposite the angle C (which is also the base of the triangle ABC).

Since you know the measure of two sides and the measure of the angle between them, you can use the Law of Cosine to find the measure of the side opposite to that angle.

So, c^2 = a^2 + b^2 - 2abcosC

a = b = 5 in, measure of the angle C = 120 degrees

c^2 = 5^2 + 5^2 - (2)(5)(5)(cos 120 degrees) = 75

c = square root of 75

c = 8. 660254038

Thus the measure length of the base is approximately 8.7 in.

(Make sure to put your calculator on degree mode).

You can use also the Pythagorean theorem to find the base, by using the property: In a right triangle the side opposite the angle whose measure is 30 degrees, is equal to one half of the measure length of the hypotenuse, but this method is much more longer.

Let's do this.

From the vertex C, draw the perpendicular CD to the base AB. This perpendicular cuts the base into two congruent parts.

In the triangle ACD the hypotenuse AC has a length of 5, so CD, which is opposite the base angle A (an 30 degrees angle) has a length 5/2 = 2.5.

From the Pythagorean theorem

AD^2 = AC^2 - CD^2

AD^2 = (5)^2 - (2.5)^2 = 25 - 6.25 = 18.75

AD = square root of 18.75

AD is approximately 4.33.

Since base AB has a length that equals twice of the length of AD, we can find that the length of AB is 2 x 4.33 = 8.66, or approximately 8.7 in.