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In △ ABC, the opposite sides of angles a, B and C are a, B and C respectively, and Tanc = 3 7． (I) find the value of COSC; (II) if CB• CA=5 2, and a + B = 9, find the length of C
(I) ∵ Tanc = 37,
In the triangle ABC, the angles a, B, C are the root 7 of a B C. Tanc = 3 times, respectively
1/(cosC)^2=(tanC)^2+1=（3√7)^2+1=64
cosC=1/8
In the triangle ABC, BC = 1, B = π / 3, and the area of the triangle is the root 3?
S triangle ABC = 1 / 2 * AB * BC * SINB = √ 3 / 4 AB = √ 3, that is ab = 4 as ad ⊥ BC intersection BC extension line at D, ad = 2 √ 3, BD = 1 / 2Ab = 2, i.e. CD = 1, so Tanc = ad / CD = - 2 √ 3
In the triangle ABC, the angle a is equal to 30 degrees, AB is equal to 4, AC is equal to 2, and the root sign is 3. Find the largest angle of BC triangle It's better to write all the formulas for solving triangles
Cosine theorem: a? = B? + C? - 2BC * cosa
It can be concluded that a 2 = 12 + 16-16 times root number 3 * (root number 3 / 2)
a²=16
A=4
So: ab = BC = 4, the triangle ABC is an isosceles triangle, we get: ∠ BAC = ∠ BCA = 30
Because the sum of the inner angles of the triangle is 180 °, so ∠ ABC = 180 ° - ∠ BAC - ∠ BCA = 120 °
Therefore: BC = 4, the maximum angle of the triangle is 120 degrees
It is known that in the triangle ABC, the angle B is equal to 45 degrees, the angle c is equal to 30 degrees, BC = 3 plus 3 radical signs 3, then the length of AB is?
Angle a equals 180-45-30 = 105 degrees
Sine theorem 3 + 3 radical sign 3 / Sina = AB / sin30
So AB is equal to three roots and two
Shuguang middle school has a triangle shaped flower bed ABC. Now we have directly measured ∠ a = 30 °, AC = 4cm, BC = 3cm, what is the area
If the height of the triangle from C intersects with ab at point D, then ad = 1 / 2 * AC = 2 (in a right triangle, the right angle of 30 degree angle is equal to half of the hypotenuse)
Shuguang middle school has a triangle shaped flower bed ABC Now we can directly measure ∠ a = 30 °, AC = 40 meters, BC = 25 meters. Please work out the area of this flower bed
In RT △ ACD, the side CD of 30 ° angle pair is equal to half of AC, that is, 40 ￣ 2 = 20ad ^ 2 = AC ^ 2 - CD ^ 2. In RT △ BCD, BD ^ 2 = BC ^ 2 - CD ^ 2 is BD = 15, so the area is s △ ABC = (15 + 20 times root 3) × 20 ￣ 2 = 150 + 2
As shown in the picture, Yingxi middle school has a triangular flower bed. Now it is measured to be ∠ a = 45 ° and BC = 5m. The height on AC is 4m. Can you work out the area of this flower bed?
∵∠A=45，BD⊥AC，
∴AD=BD=4m，
∴DC=
52−42=3．
/ / AC = 4 + 3 = 7 m,
The area of this flower bed is 7 × 4 △ 2 = 14 square meters
If ABC is known, then the root angle of AB is equal to 30
In the RT triangle ABC, the angle c = 90 degrees, the angle a = 30 degrees, and ab = a
Then BC = AB / 2 = A / 2
According to Pythagorean theorem
AC = radical (AB 2 - BC 2)
=2-cent root 3A
So AC: ab = 2, root 3
In the RT triangle ABC, ∠ C = 90 °, given a = 6, B = 2, radical 3, find ∠ B, C
In a right triangle, a ^ 2 + B ^ 2 = C ^ 2, so C = 6 ^ 2 + 2 ^ 2 = 2 √ 10,
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