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3 years ago

Cho tam giác ABC có góc B=45 độ, góc C=30 độ.Đường trung trực của cạnh BC cắt cạnh ÁC tại D.Số đo góc ABD là???

Mathematics

1 answer:

Andre45 [30]3 years ago

4 0

Answer:

Question in english please?

Step-by-step explanation:

I might be able to answer if you can put it in english.

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Can someone help please

Trava [24]

9514 1404 393

Answer:

16.4

Step-by-step explanation:

The law of cosines is useful here. It tells you ...

b^2 = a^2 + c^2 -2ac·cos(B)

b^2 = 22^2 +10^2 -2·22·10·cos(44°)

b^2 ≈ 267.49

b ≈ √267.49 ≈ 16.35514

b ≈ 16.4

4 0

3 years ago

Help pls ty in advance

svetoff [14.1K]

$\bold{\huge{\underline{ Solution\: 2 }}}$

The name of the circle is IJKL
The name of central Angle
$\sf{\underline{= {\angle} IHJ }}$
The name of the secant = KL

${\bigstar}$ Circle :- It is a round figures, which has no end points.

${\bigstar}$ Central angle :- It the angle ,which lie in the middle of the circle.

Here, H is the mid point of the circle .So, Angle IHJ is the central angle.

${\bigstar}$ Secant :- A line drawn in circle in such a way that it intersect the circle from two distinct points. Then, It is called secant .

Here, KL is the line which acts as a secant because it intersect the circle from point K and point L.

$\bold{\huge{\underline{ Solution\: 3 }}}$

Semicircle = FGH and FIH
Minor Arc :- FG or GH
Major Arc :- FHG or GFH

${\bigstar}$ Semicircle :- Semicircle is nothing but the half of the circle.

Here, In second diagram 

FGH and FIH are acting as two semicircles which simulatineously forming one circle.

${\bigstar}$ Minor Arc :- Minor Arc is nothing but the smaller arc of the circle and it is always smaller than the half of the circle that is semicircle.

Here, 

FG or GH are the minor arc of the given circle.

${\bigstar}$ Major Arc :- Major Arc is nothing but the largest arc of the circle and it is always larger than the half of the circle that is semicircle.

Here, 

FHG or GFH is the major arc as it is larger than the semicircle of the given circle .

7 0

2 years ago

An Archer shoots an arrow horizontally at 250 feet per second. The bullseye on the target and the arrow are initially at the sam

Genrish500 [490]

Answer:

Step-by-step explanation:

Using the formula for finding range in projectile, Since range is the distance covered in the horizontal direction;

Range $R = U\sqrt{\frac{H}{g} }$

U is the velocity of the arrow

H is the maximum height reached = distance below the bullseye reached by the arrow.

R is the horizontal distance covered i.e the distance of the target from the archer.

g is the acceleration due to gravity.

Given R = 60ft, U = 250ft/s, g = 32ft/s H = ?

On substitution,

$60 = 250\sqrt{\frac{H}{32}} \\\frac{60}{250} = \sqrt{\frac{H}{32}}\\\frac{6}{25} = \sqrt{\frac{H}{32}$

Squaring both sides we have;

$(\frac{6}{25} )^{2} = (\sqrt{\frac{H}{32} } )^{2} \\\frac{36}{625} = \frac{H}{32} \\625H = 36*32\\H = \frac{36*32}{625} \\H = 1.84feet$

The arrow will hit the target 1.84feet below the bullseye.

5 0

3 years ago

What are all the factor strings of 840?

nexus9112 [7]

$840|2\\420|2\\210|2\\105|5\\.\ 21|3\\.\ \ 7|7\\.\ \ 1|\\\\840=2\times2\times2\times5\times3\times7=2^3\times3\times5\times7$

6 0

3 years ago

What is equivalent to 5(9q + 7)

Andrej [43]

Answer:

45q + 35

Step-by-step explanation:

8 0

3 years ago

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