In triangle ABC, edge B = 2, diagonal B = 30 degrees, find the maximum area of triangle?

In triangle ABC, edge B = 2, diagonal B = 30 degrees, find the maximum area of triangle?

B ^ 2 = a ^ 2 + C ^ 2 - 2accosb = a ^ 2 + C ^ 2 - AC radical (3) = 4
Area s = 1 / 2 AC SINB = AC / 4
Maximum = 2 + radical (3)

It is known that in a right triangle, if the length of the larger right side is 30 and the cosine of this side is 8 / 17, then the perimeter of the triangle is and the area is

From the definition of cosine, we know that 8 / 17 is the ratio of the other straight angle side to the hypotenuse side, which can be higher than the other right angle side, which is 8x and the hypotenuse side is 17x. From the Pythagorean theorem, we know that 30 ^ 2 = (17x) ^ 2 - (8x) ^ 2. By solving this equation, we get x = 2, so the length of the other right angle side is 16 and the hypotenuse side is 34,
So the perimeter of the triangle is 80; the area is 240 (square units)

If two pieces of paper with width of 1 are crossed and overlapped together, and their intersection angle is α, the area of their overlapping part (the shaded part in the figure) is ()
A. 1sinαB. 1cosαC. sinαD. 1

As shown in the figure on the right: make AE ⊥ BC through a, AF ⊥ CD in F, perpendicular foot is e, F, ∩ AEB = ∠ AFD = 90 °, ∥ ad ∥ CB, ab ∥ CD, ∥ quadrilateral ABCD is parallelogram, ∥ strip width is 1, ∥ AE = AF = 1, in △ Abe and △ ADF, ∥ Abe = ADF = α ∥ AEB = AFD = 90 ° AE = AF, ≌ Abe ≌ ADF (AAS), ∥ AB = ad, ∥ quadrilateral ABCD is diamond The area of AB = sin α, BC = AB = 1sin α, and the overlapped part (the shadow part in the figure) is BC × AE = 1 × 1sin α = 1sin α, so a

The area of a triangular wheat field is 1620 square meters

Let's set the height of this field as H meters
36h÷2=1620
36h÷2x2=1620x2
36h=3240
36h÷36=3240÷36
h=90
A: the height of this land is 90 meters

In the triangle ABC, the angle c is a right angle, AC = 16 cm, BC = 12 cm. If the length EF of the rectangle cdef is twice the width De, the area of the rectangle can be calculated
Point E is on AB, point D is on AC, and point F is on BC.

Let the short side be a, then the total area is:
(12-a)*2a/2 + 2a*a + a* (16-2a)/2 =12*16/2 ==>
12a-a^2 +2a^2 + 8a -a^a =96 ==>
12a+8a =96
a=4.8
Area = 4.8 * 4.8 * 2 = 46.08

In △ ABC, if a = 9, B = 10, C = 12, then the shape of △ ABC is______ ．

∵ C = 12 is the largest edge, ∵ angle c is the largest angle. According to the cosine theorem, it is obtained that COSC = A2 + B2 − c22ab = 81 + 100 − 1442 × 9 × 10 ＞ 0 ∵ C ∈ (0, π), ∵ angle c is an acute angle, so a and B are also acute angles, so △ ABC is an acute triangle, so the answer is: an acute triangle

In △ ABC, if a = 9, B = 10, C = 12, then the shape of △ ABC is______ ．

∵ C = 12 is the largest edge, ∵ angle c is the largest angle. According to the cosine theorem, it is obtained that COSC = A2 + B2 − c22ab = 81 + 100 − 1442 × 9 × 10 ＞ 0 ∵ C ∈ (0, π), ∵ angle c is an acute angle, so a and B are also acute angles, so △ ABC is an acute triangle, so the answer is: an acute triangle

In △ ABC, if a = 9, B = 10, C = 12, then the shape of △ ABC is______ ．

∵ C = 12 is the largest edge, ∵ angle c is the largest angle. According to the cosine theorem, it is obtained that COSC = A2 + B2 − c22ab = 81 + 100 − 1442 × 9 × 10 ＞ 0 ∵ C ∈ (0, π), ∵ angle c is an acute angle, so a and B are also acute angles, so △ ABC is an acute triangle, so the answer is: an acute triangle

In the triangle ABC, a = 7, B = 10, C = 6, shape?
RT

According to the cosine theorem
a^2+c^2-b^2

In △ ABC, if the lengths of its three sides are 9, 40 and 41 respectively, then the area of the triangle

∵9^2+40^2=1681=41^2,
And Δ ABC is a right triangle,
∴SΔABC=1/2×9×40=180.