In an isosceles triangle, if a base angle has a measure of 43 degrees, then the vertex angle must have a measure of 80 degrees as well.
Nope. The two base angles are equal, making them add to 86° The vertex angle n=must be (180-86)° = 94°
Triangle ABC is an isosceles triangle. AB is the longest side with length 4x + 4. BC = 8x + 3, and CA = 7x + 8. Find AB
Doing a piggyback, eh?
If AB is the longest side, that must mean that the other two sides are equal: AC=BC
8x+3 = 7x+8
x = 5
AC=BC=43
so, AB = 4x+4 = 24
Hey! I thought AB was the longest. In fact, if x>0, then 4x+4 > 8x+3 only if x < 1/4, so the +4 outdoes the +3.
Looks like for any reasonable x, 8x+3 will be the largest value.
22 degrees an isosceles triangle is shown what is the measure of
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To determine the measure of the vertex angle in an isosceles triangle, we need to know the measure of one of the base angles. In an isosceles triangle, the base angles are congruent, which means they have the same measure.
Given that one of the base angles measures 43 degrees, we can conclude that the other base angle also measures 43 degrees. This is because in an isosceles triangle, the base angles are equal.
To find the measure of the vertex angle, we need to subtract the sum of the base angles from 180 degrees, because the sum of interior angles in a triangle is always 180 degrees.
Let's calculate it:
Measure of vertex angle = 180 degrees - (43 degrees + 43 degrees)
= 180 degrees - 86 degrees
= 94 degrees
Therefore, the vertex angle in this isosceles triangle measures 94 degrees, not 80 degrees.