Given that an interior angle of an isosceles triangle is 80 degrees, then the top angle of the isosceles triangle is () degrees

Given that an interior angle of an isosceles triangle is 80 degrees, then the top angle of the isosceles triangle is () degrees

If an interior angle of an isosceles triangle is known to be 80 degrees, then the top angle of the isosceles triangle is (20 or 80) degrees

It is known that the base angle of an isosceles triangle is 50 ° and what is the top angle of this isosceles triangle?

180°-50°×2
=180°-100°
=80°
A: the top angle of this isosceles triangle is 80 degrees

If the sum of the three inner angles of an isosceles triangle and an outer angle of the vertex angle is equal to 280?, then its internal angles are respectively?

If the three inner angles are 180 degrees, then the outer angle is 100 degrees, and the one tangent to the outer angle is 80 degrees, then the isosceles triangle is 180-80 = 100 degrees, 100 / 2 = 50 degrees
The three internal angles are 80.50.50 degrees

An interior angle of an isosceles triangle is equal to one fourth of the other

If the top angle is large, then a base angle is equal to 1 / 4 of the top angle
That is, one top angle is equivalent to four bottom angles
So each base angle = 180 / (4 + 1 + 1) = 30 degrees
Vertex angle = 30 × 4 = 120 degrees
If the top angle is small, then a base angle is equivalent to four top angles
So each vertex angle = 180 / (4 + 4 + 1) = 20 degrees, and the vertex angle is regarded as a multiple

A diplomatic sum of the three internal angles and the top angle of an isosceles triangle is equal to 280 ° so what are the internal angles of the isosceles triangle?

The sum of the interior angles of the triangle is 180 degrees, so the outer angle of the top angle is 280-180 = 100 degrees
The top angle is 180-100 = 80 degrees
Both base angles are (180-80) / 2 = 50 degrees

If the angle between the height on one waist and the height on the other is 45 degrees, then the top angle of the isosceles triangle is ()

As shown in the figure:
CD is the height on the side of AB, and the angle between Cd and AC on the other side is 45 degrees,
That is to say, in RT △ ADC, ∠ DCA = 45 & ordm;, then ∠ a = 45 & ordm;

If the angle between the height of one waist and the other is 45 degrees, the base angle of isosceles triangle is () A. 67° B. 67.5° C. 22.5° D. 5 ° or 22

There are two situations;
(1) As shown in the figure, when △ ABC is an acute triangle, BD ⊥ AC is at D,
Then ∠ ADB = 90 °,
It is known that ∠ abd = 45 °,
∴∠A=90°-45°=45°，
∵AB=AC，
∴∠ABC=∠C=1
2×（180°-45°）=67.5°；
(2) As shown in the figure, when △ EFG is an obtuse triangle, FH ⊥ eg is at h,
Then ∠ fhe = 90 °,
It is known that ∠ HFE = 45 °,
∴∠HEF=90°-45°=45°，
∴∠FEG=180°-45°=135°，
∵EF=EG，
∴∠EFG=∠G=1
2×（180°-135°）=22.5°，
According to (1) (2), the base angle of isosceles triangle is 67.5 ° or 22.5 °
Therefore, D

If the angle between the height on one waist and the other waist of an isosceles triangle is 45 degrees, what is the base angle of the isosceles triangle? A: 67.5 degrees, B: 22.5 degrees C: 22.5 degrees or 67.5 degrees

C
If the acute angle triangle
Top angle = 45 base angle = (180-45) divided by 2 = 67.5
If obtuse triangle
Top angle = 180-45 = 135 bottom angle = (180-135) divided by 2 = 22.5

The angle between the height of one waist and the base of an isosceles triangle is equal to () A. Half of the top angle B. Half of the bottom corner C. 90 ° minus half of the vertex angle D. 90 ° minus half of the base angle

In △ ABC, ∵ AB = AC, BD is high,
∴∠ABC=∠C=180−∠A
Two
In RT △ BDC, ∠ CBD = 90 ° - ∠ C = 90 ° - 180 − a
2=∠A
2．
Therefore, a

If the vertex angle of an isosceles triangle is α, then the angle between the height of one waist and the base of an isosceles triangle is equal to () A. α Two B. 90°+α Two C. 90°-α Two D. 90°+α

According to the meaning of the title, base angle = 1
2（180°-α）=90°-α
2，
The included angle is 90 ° - (90 ° - α)
2）=α
2．
Therefore, a