The triangle ABC is an equilateral triangle, O is a point in the triangle ABC, OA = 5. Ob = 4. OC = 3. Find the degree of angle BOC

The triangle ABC is an equilateral triangle, O is a point in the triangle ABC, OA = 5. Ob = 4. OC = 3. Find the degree of angle BOC

Rotate △ BOC around C so that BC and AC coincide, O falls at o 'to obtain △ ACO', and connect oo`
Then OC = OC '∠ OCO' = 60 °
∴OO`=3 ∠OO`C=60°
In △ aoo ', OO' = 3 Ao = 5 Ao '= 4
∴∠AO`O=90°
∴∠AO`C=90°+60°=150°
∴∠B0C=150°

As shown in the figure, we know that in the Pentagon ABCDE, AE is parallel to CD, angle a = 100 degrees, angle B = 130 degrees, and find the degree of angle C Angle a = 100 degrees, angle B = 130 degrees~

The sum of internal angles of Pentagon is 540 degrees, angle a is 100 degrees and angle B is 130 degrees. Because AE and CD are parallel, angle E and angle d together are 180 degrees, and angle c = 540-100-130-180 = 30 degrees

As shown in the figure, in the Pentagon ABCDE, AE ∥ BC, angle d = 100 ° angle E ratio angle c = 7:6, calculate the degree of angle E and angle C

Connecting CE ∵ AE / / BC,  AEC  BCE = 180 ° in △ CDE, ? CED  ECD = 180 ° - ﹤ d = 180 ° - 100 ° = 80 °  AEC ? BCE ｛ CED ｛ ECD = 180 ° + 80 ° = 260 ° i.e., ﹤ AED ﹤ BCD = 260 ° and  AED: ﹤ BCD = 7:6, ﹤ AED = 7K

As shown in the figure, in the Pentagon ABCDE, AE ‖ CD, ∠ a = 107 ° and ∠ ABC = 121 ° are used to calculate the degree of ∠ C

BF ∥ AE is made on the right side of B through point B
∵BF∥AE，∠A=107°，
∴∠ABF=180°-107°=73°，
∵∠B=121°，
∴∠FBC=121°-∠ABF=48°，
AE ∥ CD, BF ∥ AE,
∴BF∥CD，
∴∠C=180°-∠FBC=132°．

In the Pentagon ABCDE, AE is parallel to CD; angle a = 130 ° angle c = 110 ° try to find the degree of angle B

The sum of internal angles of Pentagon is equal to (5-2) * 180 degrees = 540 degrees. According to the knowledge of parallel lines, the angle e + angle d = 180 degrees, so the angle B = 120 degrees

As shown in the figure, in the Pentagon ABCDE, AE / / CD, angle a = 107 ° and angle B = 121 ° to find the degree of angle C

AE / / CD, so the sum of angles E and D is 180 degrees, the sum of internal angles of Pentagon is 540, a = 107, B = 121, e + D = 180,
So C = 132 degrees

As shown in the figure, given that the quadrilateral ABCD is inscribed in ⊙ o, ∠ BOD = 80 ° and the degrees of ∠ bad and ∠ BCD are obtained

∵∠BOD=80°，∴∠BAD=40°．
Also ∵ ABCD is the inscribed quadrilateral of a circle,
∴∠BAD+∠BCD=180°，
∴∠BCD=140°．

As shown in the figure, ∠ BOD = 160 ° then the degree of ∠ bad? ∠ BCD is? Circle inscribed quadrilateral is isosceles trapezoid, Bo do is radius in trapezoid

The degree of ∠ BCD is half of 160 ° i.e. 80 ° because the circle angle opposite to the same arc is half of the center angle
The degree of bad is 100 ° because it is a circle inscribed quadrilateral and diagonally complementary

As shown in the figure, a, B, C, D are the four points on circle O, and the angle BCD = 100 °, find the size of the angle BOD (the center angle of the circle to which the arc BCD is opposite) and lbad?

The circumference angle is half of the center angle, so ∠ BOD = 2 ∠ BCD, so the large angle ∠ BOD = 200 degrees. Because the circumference angle is 360 degrees, the small angle ∠ BOD = 160 degrees, so ∠ bad = 80 degrees
You can solve it in another way

As shown in the figure, the quadrilateral ABCD is inscribed in ⊙ o, if ∠ BOD = 140 °, then ∠ BCD=______ .

∵∠BOD=140°，
∴∠A=1
2∠BOD=70°，
∴∠BCD=180°-∠A=110°．
So the answer is 110 degrees