You need trig to calculate such angles; it has it's time .
The sides of the triangle are all surface diagonals of a cube, digonals of congruent/equal squares .
Find first the indicated 2 right angles .
Take care of the 2 isosceles triangles .
Wind β which is equal ...
The angle between purple line segments should be 90° because ...
2 triangles should have a total of 360°.
Angles whose sides are perpendicular are equal.
Have please a look at ry1 .
The line segments at top and bottom are parallel .
The sum of angles to the right (180°-70°) is supposed to be equal to the left [unkown angle + (180°-130°)] .
Bisector divides the unknown angle into 2 and the sum of inner angles of a triangle ...
There are 2 possibilities (note that these are consistent with each othar):
1- The sum of the outer angles of the triangle is 360° so find the 3rd outer ...
2- The supplementaries of the given angles are 30° and 60° so use the sum of the inner angles of a triangle ...
Isosceles triangles do have equal base angles ... .
There are 2 triangles and their respective three angles always add to 180° .
1- m(‹O) = 180° - [m(‹B)+m(‹C)]/2 .
2- m(‹A) = 180° - [m(‹B)+m(‹C)]
2 ==> m(‹A/2) = 90° - [m(‹B)+m(‹C)]/2
1&2 ==> m(‹O) = 180° - [90° - m(‹A/2)] ...
Divide the quadrilateral into 2 triangles and the sum of inner angles of each ...
150° - (180° - 2·a) = 2·a - 30°
2 · (2·a - 30°) + 90° - a = 180° .
Have please a look at ry1.
3·a + 60° = 180° .
Inscribed angles subtending the same arc are equal. Remember that one of the base angles of the right triangle was 50° and it should have a vertex angle
of ...
A second way of solving: The inscribed angle subtending the same common arc with a central angle is half of the central.
Just like the inscribed angle ... .
The inscribed angle is 120° , central angle/chord will be 240° .
The rest 120° of the chord is twice the sum of the both tangent-chord angles.
The sum of the inner angles of a quadrilateral is 360° .
18·a = 180° .
The green line segments have equal lengths, take please care .
Have please a look at ry1. If the 3 angles of the quadrilateral at right is found, 4th will be obtained by subtracting their sum from 360° .
One of the angles of the quadrilateral is right angle .
Yellow triangle is an isosceles one, its base angle is (180° - 155°)/2 .
One of the angles of the quadrilateral is tangent-chord angle .
The sum of the 2 upper/vertex angles of the quadrilateral is 360° - 220° = 140° .
The angle that completes this to 2·180° = 360° is 220°, the sum of the 2 angles having bisectors .
This means the sum of the base angles of the yellow tringle is 110° .
The sum of the base angles of the triangle is [360° - (a+b)]/2 .
The vertical angles of (light) yellow triangles have to be (66°+50°)/2 = 58° .
Have please a look at ry1 .
Connect points A and be to have a 7 sided polygon whose sum of inner angles is (7-2)·180°=900° .
The sum of the angles given is 775° .
Tangent (chord) angle is half of central .
Inscribed angle is half of central .
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