An exterior angle of an isosceles triangle has measure 150°. Find two possible sets of measures for the angles of the triangle.
1 answer:
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Answer: The answers is alternate interior angles.
Step-by-step explanation: First of all, the questions marks given in the figure are renamed in the attached figure as (a), (b), (c) and (d).
For (a): Since AC is parallel to A'C' and A'D is a transversal for these two parallel lines, so, ∠CDB' = ∠B'A'C', because these are alternate interior angles.
For (b): Since BC is parallel to B'C' and A'B' is a transversal, so ∠BEB' = ∠A'B'C', because these are alternate interior angles.
For (c): Since AB is parallel to A'B' and AD is a transversal, so ∠BAC = ∠CDB', because these are alternate interior angles.
For (d): Since AB is parallel to A'B' and BE is a transversal, so ∠ABC = ∠BEB', because these are alternate interior angles.
Thus, all the questions marks are the reasons that the given angles are equal because they are alternate interior angles.
Let z = sin(x). This means z^2 = (sin(x))^2 = sin^2(x). This allows us to go from the equation you're given to this equation: 7z^2 - 14z + 2 = -5
That turns into 7z^2 - 14z + 7 = 0 after adding 5 to both sides. Use the quadratic formula to solve for z. The only solution is z = 1 (see attached image). Since we made z = sin(x), this means sin(x) = 1. All solutions to this equation will be in the form x = (pi/2) + 2pi*n, which is the radian form of the solution set. If you need the degree form, then it would be x = 90 + 360*n
The 2pi*n (or 360*n) part ensures we get every angle coterminal to pi/2 radians (90 degrees), which captures the entire solution set.
Note: The variable n can be any integer.
