Given that the lengths of both sides of the triangle are 5 and 7 respectively, the value range of the center line length x of the third side is () A. 2＜x＜12 B. 5＜x＜7 C. 1＜x＜6 D. Unable to determine

Given that the lengths of both sides of the triangle are 5 and 7 respectively, the value range of the center line length x of the third side is () A. 2＜x＜12 B. 5＜x＜7 C. 1＜x＜6 D. Unable to determine

Extend ad to e so that ad = De,
As shown in the figure, ab = 5, AC = 7,
Let BC = 2A, ad = x,
In △ BDE and △ CDA,
AD=DE
∠ADC=∠BDE
BD=CD ，
∴△BDE≌△CDA，（SAS）
∴AE=2x，BE=AC=7，
In △ Abe, be-ab ＜ AE ＜ AB + be, that is, 7-5 ＜ 2x ＜ 7 + 5,
∴1＜x＜6．
Therefore, C

The three sides of DOC triangle are 3 and 5 respectively. What is the value range of the center line of the third side?

two

(2011. Zhengzhou three mode) in Delta ABC, tanA = 1 2，cosB＝3 ten 10, then the value of Tanc is () A. -1 B. 1 C. three D. 2

sinB=
1−cos2B=
ten
10，tanB=sinB
cosB=1
three
tanC=tan（180°-A-B）=-tan（A+B）=-tanA+tanB
1−tanAtanB=-1
So choose a

In triangle ABC, ∠ a, ∠ B and ∠ C correspond to edges a, B and C respectively, and it is known that (a + B + C) (a-b + C) = 3aC, Tana + Tanc = 3 + radical 3 AB side height is 4 times root 3 to find the degree of each angle and the length of three sides

(a + B + C) (a-b + C) = 3aC, (a + C) ^ 2-B ^ 2 = 3aC, a ^ 2 + C ^ 2-B ^ 2 = AC, CoSb = (a ^ 2 + C ^ 2-B ^ 2) / 2Ac = 1 / 2, B = π / 3 = 60 degrees. Tan (a + C) = - tanb = - √ 3, i.e. (Tana + Tanc) / (1-tanatanc) = (3 + √ 3) / (1-tanatanc) = - √ 3 tanatanc = 2 + √ 3. (1) Tana + Tanc = 3 + √ 3. (2)

Known in △ ABC, a= 5，b= 15, a = 30 °, find C

SINB = bsina from sine theorem
a=
15sin30°
5=
three
2，
And ∵ b > A,
‡ B ＞ a, so B = 60 ° or 120 °
(1) When B = 60 °, C = 90 °
According to Pythagorean theorem:
∴c=
a2+b2=2
5，
(2) When B = 120 °, C = a = 30 °
∴c=a=
5，
To sum up: C=
5 or 2
five
So the answer is: C=
5 or 2
five

In triangle ABC, a, B and C are the opposite sides of ∠ a, B and C respectively. Given that a = root 3, B = 3 and ∠ C = 30 degrees, what is ∠ a equal to

Cosine theorem COSC = (a) ²+ b ²- c ²)/ 2Ab substitute a = √ 3, B = 3, C = 30 ° into the above formula, √ 3 / 2 = (3 + 9-c) ²)/ 6 √ 3 solution, C = √ 3 = A. therefore, triangle ABC is isosceles triangle, ∠ C = ∠ a = 30 ° [in addition] sine theorem C / sinc = B / SINB = A / sina will a = C = √ 3, C = 30 °