In triangle ABC, angle c is equal to 90 degrees, angle a is equal to 15 degrees, and BC is equal to 4. Find the area of triangle ABC The figure is a single triangle

In triangle ABC, angle c is equal to 90 degrees, angle a is equal to 15 degrees, and BC is equal to 4. Find the area of triangle ABC The figure is a single triangle

Make de ⊥ AB crossing AC at e through the midpoint D of AB, ≓ De is the vertical line of AB, ⊙ AE = be, ∠ EBA = ∠ a = 15 °, ⊆ BEC = 15 ° + 15 ° = 30 °, (outer angle = sum of two non adjacent inner angles) and ∠ C = 90 °, ⊙ BC = be / 2, be = 2BC = 2 * 4 = 8 = aece = √ (be ²- BC ²)= √(8 ²- four ²)= 4√3AC=A...

As shown in Figure 1, in RT triangle ABC, angle c = 90 degrees, angle a = 15 degrees, BC = 1, calculate the area of triangle ABC

Make the vertical bisector of AB, intersect AB at point D, intersect AC at point E, and connect be
Then EA = EB
∴∠EBA=∠A=15°
∴∠BEC=30°
∵BC=1
∴BE=AE=2,CE=√3
∴AC=2+√3
∴S△ABC=1/2(2+√3)*1=(2+√3)/2

In triangle ABC, ab = 15, BC = 14, AC = 13 find the area of triangle ABC Have an answer by tomorrow! Come on!

Maybe you haven't learned this knowledge. You use Helen's theorem. It is to find the area according to the known three side lengths of triangles
S = P (P-A) (P-B) (P-C), and then open the root of the obtained S. in the formula, a, B and C refer to the three sides of the triangle respectively
In the formula, P = (a + B + C) / 2 = the perimeter of the triangle divided by 2
Now we will substitute this formula into the problem to calculate
Step 1: first calculate P = (15 + 14 + 13) / 2 = 21
Step 2: calculate s = 21 * (21-15) * (21-14) * (21-13)
s=21*6*7*8
s=7056
Step 3: open the calculated number to the root sign, and the number under the root sign is 7056 = 84
That's it

In triangle ABC, B = 5, C = 6, triangle area = (15 times root sign 3) / 2, find a? .

In triangle ABC, B = 5, C = 6, triangle area = (15 times root sign 3) / 2, find a? Area formula s = BC * sin a = 5 * 6 * Sina = = (15 times root sign 3) / 2, so sin a = root sign 3 / 2cosa = 1 / 2 or - 1 / 2 according to cosine theorem a ^ 2 = B ^ 2 + C ^ 2 - 2 * BC * cosa = 31 or 91, so a = root sign 31

In triangle ABC, the triangle area is equal to 15 times the root sign 3, a + B + C = 30 ∠ a + ∠ C is equal to half of ∠ B. what is the length of three sides of the triangle a, B and C?

Provide you with ideas: from ∠ a + ∠ C is equal to half of ∠ B, combined with the sum of triangles is 180 °, know that ∠ B = 120 °, apply sine theorem and triangle area is equal to 15 times root sign 3, get 1 / 2 * a * c * sin ∠ B = 180 ° solution, get a * C = 60 (1) apply cosine theorem: cos ∠ B = (a ^ 2 + B ^ 2-C ^ 2) / 2Ac solution, get B ^ 2-A ^ 2 -

In triangle ABC, a, B and C are the opposite sides of ∠ a, ∠ B and ∠ C respectively. If 2B = a + C and ∠ B = 30 °, the area of triangle ABC is 1 / 2, find B

From the positive area formula: S = (1 / 2) acsinb = 1 / 2
Obtained: AC = 2
(2
Cosine justification:
b^2=a^2+c^2-2ac*cosB.
Because: A ^ 2 + C ^ 2 = (a + C) ^ 2-2ac
Therefore, the above formula is:
b^2=a^2+c^2-2ac*cosB
=(a + C) ^ 2-2ac-2ac * CoSb is substituted into the known conditions to obtain
B ^ 2 = (2b) ^ 2-2 * 2-2 * 2 (radical 3) / 2
Sorted: 3B ^ 2 = 4 + 2 * (root sign 3)
B ^ 2 = [4 + 2 * (root 3)] / 3=
=(1 / 3) * [(root 3) + 1] ^ 2
So B = (1 / root 3) * [(root 3) + 1]
=1 + (root 3) / 3