As shown in the figure, △ ABC ≌ △ ade, the corresponding vertex of point B is point D. if ∠ bad = 100 °, CAE = 40 °, calculate the degree of ∠ BAC

As shown in the figure, △ ABC ≌ △ ade, the corresponding vertex of point B is point D. if ∠ bad = 100 °, CAE = 40 °, calculate the degree of ∠ BAC

∵△ABC≌△ADE，
∴∠BAC=∠DAE，
∴∠BAC-∠CAE=∠DAE-∠CAE，
That is ∠ BAE = ∠ DAC,
∵∠BAD=100°，∠CAE=40°，
∴∠BAE=1
2（∠BAD-∠CAE）=1
2（100°-40°）=30°，
∴∠BAC=∠BAE+∠CAE=30°+40°=70°．

In the triangle ABC, the angle BAC is equal to 100 degrees, BD bisection angle ABC intersects AC with D, AB equals AC. it is proved that BD + ad = BC

Intercept be = BD on BC,
∵ BD bisection angle ABC intersects AC with D
∴∠DBC=20°
∴∠BED=80°
And ∵ AB = AC, ∵ BAC = 100 
∴∠C=40°
∴∠CDE=40°
∴CE=DE
Through D, make DM ⊥ BC to m, make DN ⊥ Ba to Ba extension line to n,
Then DM = DN, ∠ bed = ∠ Dan = 80 °
∴△DAN≌△DEM
∴DE=DA
∴CE=DA
∴BC=BE+CE=BD+AD

As shown in the figure, ∠ BAC = 100 ° in △ ABC, DF and eg are the vertical bisectors of AB and AC respectively, then ∠ DAE is equal to______ Degree

∵ DF and eg are the vertical bisectors of AB and AC respectively
If (1) Da = dB, then ∠ B = ∠ DAF, let ∠ B = ∠ DAF = x degree
(2) EA = EC, ∠ C = ∠ EAG, let ∠ C = ∠ EAG = y degree
Because ∠ BAC = 100 °
Therefore, x + y + ∠ DAE = 100 °
According to the triangle interior angle sum theorem, x + y + X + y + ∠ DAE = 180 degrees
It is found that ∠ DAE = 20 °

In the triangle ABC, angle B = 1 / 3, angle a = 1 / 4 angle C. find the degree of angle C

Let B = K. then a = 3k, C = 4K
If a + B + C = 180, then 8K = 180, k = 22.5, C = 90

In the triangle ABC, the angle A and the angle B are equal to the angle C. the triangle must be () and the angle must be adopted!

Acute angle

A = 16, B = 5, C = 19. Find the area of triangle ABC As the title

cosA=(b^2+c^2-a^2)/2bc=(25+361-256)/2*5*19=13/19
(sinA)^2+(cosA)^2=1
And a is the inner angle of the triangle
So Sina 〉 0
So Sina = 8 √ 3 / 19
S=bcsinA/2=20√3

In △ ABC, given a = 3, B = 2, C = √ 19, find the size and area of this triangle

cosC=(3^2+2^2-19)/(2*3*2)=-1/2
sinC=√3/2,C=120
SABC=1/2absinC=3√2/2

In the triangle ABC, given that C = root 3, B = 1, B = 30 degrees, find the values of angles C and a, and find the area of the triangle

By CoSb = (C ^ 2 + A ^ 2-B ^ 2) / 2Ac
A = 2, or 1
When a = 2, a = 90, C = 60, area = √ 3x1x1 / 2 = √ 3 / 2
When a = 1, a = 30, C = 120. Area = 1 / 2xsinbxac = 1 / 2xsin30x √ 3x1 = √ 3 / 4

In the triangle ABC, we know a = root 3. B = 1. B = 30 degrees. Find a and triangle area

When C = 1, a = 120 degrees, s = root 3 / 4. When C = 2, a = 60 degrees, s = root 3 / 2

Let a, B, C be the three sides of the triangle ABC with an area of 12 3, BC = 48, B-C = 2, then a =___ .

When a = π 3, from the cosine theorem A2 = B2 + c2-2bccosa = 64 + 36-96 × 12 = 52,  a = 213, when a = 2 π 3, we can get a = 237