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In the triangle ABC, if edge B + C = radical 2 + 1, angle c = 45 degrees, angle B = 30 degrees, what are the values of edges B and C
B = 1, C = radical 2
Make a perpendicular line from point a to BC to intersect BC and D, then ad = B divided by root 2 is equal to 0.5c, then B divided by root 2 equals to 0.5c
Then there is B + C = radical 2 + 1
B = 1, C = radical 2
In the triangle ABC, ABC is the opposite side of the angle ABC respectively, if Tana = 3, COSC = the root of 5, if C = 4 angle B45 degree s triangle area=
In this paper, the sine of the angle a is 3 times the root 10, and the sine of the angle c is 2 times the root 5
As shown in the figure, ∠ BAC = 60 ° and ab = 2Ac. Point P is in △ ABC, and pa= 3, Pb = 5, PC = 2, calculate the area of △ ABC
As shown in the figure, make △ ABQ such that ∠ qAB = ∠ PAC, ∠ ABQ = ∠ ACP, then △ ABQ ∽ △ ACP
∵AB=2AC，
The similarity ratio of △ ABQ and △ ACP was 2
∴AQ=2AP=2
3，BQ=2CP=4，
∠QAP=∠QAB+∠BAP=∠PAC+∠BAP=∠BAC=60°．
According to AQ: AP = 2:1, ∠ Apq = 90 ° and PQ=
3AP=3，
﹣ bp2 = 25 = bq2 + pq2, thus ﹣ BQP = 90 °,
After passing through point a for am ∥ PQ, extend BQ and cross am to point M,
∴AM=PQ，MQ=AP，
∴AB2=AM2+（QM+BQ）2=PQ2+（AP+BQ）2=28+8
3，
So s △ ABC = 1
2AB•ACsin60°=
Three
8AB2=6+7
Three
2=3+7
Three
2．
So the answer is: 3 + 7
Three
2．
As shown in the figure, ∠ BAC = 60 ° and ab = 2Ac. Point P is in △ ABC, and pa= 3, Pb = 5, PC = 2, calculate the area of △ ABC
As shown in the figure, make △ ABQ such that ∠ qAB = ∠ PAC, ∠ ABQ = ∠ ACP.
As shown in the figure, the side length a of the equilateral △ ABC= 25+12 3. The point P is a point in △ ABC, and pa2 + PB2 = PC2. If PC = 5, find the length of PA and Pb
With B as the center, rotate △ BAP by 60 degrees so that point a rotates to point C and point P to point Q
∵PA2+PB2=PC2
The △ PCQ is a right triangle, ∠ CQP = 90 °
∴∠CQB=150°．
BC2=CQ2+BQ2-2CQ•BQcos150°
=PA2+PB2-2PA•PB（-
Three
2）
=PC2+
3PA•PB
=25+
3PA•PB．
BC2=25+12
3．
∴PA•PB=12，
∵PA2+PB2=25，
ν PA = 3, Pb = 4 or PA = 4, Pb = 3
As shown in the figure, ∠ BAC = 60 ° and ab = 2Ac. Point P is in △ ABC, and pa= 3, Pb = 5, PC = 2, calculate the area of △ ABC
As shown in the figure, make △ ABQ such that ∠ qAB = ∠ PAC, ∠ ABQ = ∠ ACP, then △ ABQ ∽ △ ACP
∵AB=2AC，
The similarity ratio of △ ABQ and △ ACP was 2
∴AQ=2AP=2
3，BQ=2CP=4，
∠QAP=∠QAB+∠BAP=∠PAC+∠BAP=∠BAC=60°．
According to AQ: AP = 2:1, ∠ Apq = 90 ° and PQ=
3AP=3，
﹣ bp2 = 25 = bq2 + pq2, thus ﹣ BQP = 90 °,
After passing through point a for am ∥ PQ, extend BQ and cross am to point M,
∴AM=PQ，MQ=AP，
∴AB2=AM2+（QM+BQ）2=PQ2+（AP+BQ）2=28+8
3，
So s △ ABC = 1
2AB•ACsin60°=
Three
8AB2=6+7
Three
2=3+7
Three
2．
So the answer is: 3 + 7
Three
2．
As shown in the figure, ∠ BAC = 60 ° and ab = 2Ac. Point P is in △ ABC, and pa= 3, Pb = 5, PC = 2, calculate the area of △ ABC
As shown in the figure, make △ ABQ such that ∠ qAB = ∠ PAC, ∠ ABQ = ∠ ACP.
Let p be in the triangle ABC, and PA = radical 3, Pb = 5, PC = 2, angle BAC = 60 degrees. Find the area of triangle ABC (how to prove that it is a right triangle?) AB=2AC
It can be obtained by cosine theorem
Square of BC = square of AB + square of AC - 2 * AB * ac * cos angle bac
Bring in ab = 2Ac
The square of BC = 5AC - 4 * ac * 1 / 2
=The square of 3aC
BC = radical 3aC
So we can get AB squared = BC squared + AC squared
P is the interior point of the equilateral triangle ABC, PA = 4 Pb = 2 root sign 3 PC = 2. Find the area of triangle ABC Today's inner Express
Draw a picture, and then rotate counterclockwise 60 degrees around vertex a to make B reach the position of C, C to C ', P to p'
Connect p'a, p'c, PP '
PA＝P'A＝2
Angle PAP '= 60 degrees
So PP '= 2
In △ pp'c, PC = 4, p'c = double root 3, PP '= 2
So angle CPP '= 60 degrees, angle cp'p = 90 degrees
So angle APC = 60 + 60 = 120 degrees, angle APB = angle ap'c = 60 + 90 = 150 degrees
So the angle BPC = 360-120-150 = 90 degrees
The result shows that the side length of equilateral triangle BC = 2 √ 7
Area = 1 / 2 * 2 √ 7 * 2 √ 7 * √ 3 / 2 = 7 √ 3
In the triangle ABC, the angle ABC = angle c, BD is the bisector of angle ABC, and the angle BDC = 87 degrees?
BD is the bisector of angle ABC
So angle ABC = 2x angle DBC
Angle ABC = angle c = 2x angle DBC
Angle c + angle DBC + angle BDC = 180 degrees
So the angle abd = angle DBC = 31 degrees
Angle a = angle BDC - angle abd = 56 degrees
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