In ABC, ab = 7, AC = 5, BC = 3, find SINB, CoSb

In ABC, ab = 7, AC = 5, BC = 3, find SINB, CoSb

According to cosine theorem, CoSb = (7 ^ 2 + 3 ^ 2-5 ^ 2) / (2 * 7 * 5) = 33 / 72 = 11 / 24,
From the square relation, SINB = √ [1 ^ 2 - (11 / 24) ^ 2] = √ 455 / 24

In the triangle ABC, ab = AC = 13, BC = 10, find the value of tanb, CoSb, SINB

Make a vertical line of BC through point a and intersect BC and D. because AB = AC, point D is the midpoint of BC
AD²=AB²-BD² =169-25=144 AD=12
tanB=AD/BD=12/5 cosB=BD/AB=5/13 sinB=AD/AB=12/13
Complete,

In △ ABC, if AB = AC and CoSb = 15, then bcab=______ ．

Make BC side vertical line ad. ∵ AB = AC, ∵ BD = CD. If ∵ CoSb = 15, ∵ bdab = 15, then AB = 5bd, BC = 2bd, ∵ bcab = 25

In the isosceles triangle ABC, ab = AC, BD bisects ∠ ABC intersects AC with D, if ∠ CDB = 150 degrees, then ∠ C=
It's not 30. I'll give you a why

Let two equal angles be X,
Then ∠ BCD = x, ∠ DBC = x / 2
Known ∠ CDB = 150
In △ BDC
∠BCD+∠DBC+∠CDB=180
x+x/2+150=180
x*3/2=30
The solution is x = 20 = ∠ C

In the isosceles triangle ABC, AB equals AC, BD flat angle ABC intersects AC with D, angle CDB equals 150, then how many degrees does angle a equal () a.130 degrees b.140 degrees c.150 degrees d.160 degrees

B. 140 degrees
In the isosceles triangle ABC, AB equals AC, BD flat angle ABC intersects AC with D, angle CDB equals 150, then how many degrees does angle a equal () a.130 degrees b.140 degrees c.150 degrees d.160 degrees

As shown in the figure, it is known that in △ ABC, ab = AC, ∠ a = 40 °, the bisector BD of ∠ ABC intersects AC at D. find the degree of ∠ ADB and ∠ CDB

∵ AB = AC, ∵ a = 40 °, ∵ ABC = ∵ C = (180 ° - 40 °) △ 2 = 70 °, and ∵ BD is the bisector of ∵ ABC, ∵ abd = ∵ CDB = 35 °, ∵ ADB = 180 ° - (40 ° + 35 °) = 105 °. Therefore, the degree of ∵ ADB is 105 ° and the degree of ∵ CDB is 105 °

In triangle ABC
A 80 B 110 C140 D is wrong, no more choices

It is known that BD is the height of one waist of the isosceles triangle ABC, and the angle DBA = 50 ° to find the degree waist drawing of the three inner angles of △ ABC
It's better to draw the picture well, too

How to draw a picture? Find the degree of three internal angles: angle cab is 40 °, angle ACB = angle ABC is 70 °

The angle between the height of one waist and the other waist of an isosceles triangle is 50 degrees
It's a process

If the angle between the height of one waist and the other waist of an isosceles triangle is 50 degrees, then the vertex angle of the isosceles triangle is_ 40 degrees or 140 degrees___ .
90-50 = 40 degrees when high on one waist;
180 - (90-50) = 140 degrees on the extension line of one waist

It is known that BD is an isosceles triangle, and the height of one waist of ABC is less than abd = 40 degrees

Vertex angle = 90 ° - 40 ° = 50 °
or
Apex angle = 180 ° - (90 ° - 40 °) = 180 ° - 50 ° = 130 °
A vertex angle is an acute angle
A vertex angle is obtuse, which is the high BD on the outside of the triangle