in an isoselese triangle ACB, AB=AC=4.BAC=30° and CD ia ab altitude.Find BN.(explain witj diagram)
No idea where N is, so can't help there.
But BC = 2*4sin15°
BD = BC sin15°
DA = 4 cos30°
DC = 4 sin30°
To find BN in an isosceles triangle ACB, where AB = AC = 4 and BAC = 30°, we need to draw a diagram first.
Step 1: Draw a horizontal line AB. Mark point C above point B such that AB = AC = 4.
```
C
/\
/ \
/ \
/......\
/________\
A 4 B
```
Step 2: Find the altitude CD. Draw a perpendicular line from point C down to line AB, and label the point of intersection as D.
```
C
/\
/ \
/ \
/......\.
/________\
A 4 B
|
|
D
```
Step 3: Since triangle ACB is isosceles, AD is equal to BD. Split line CD in half at point N.
```
C
/\
/ \
/ n\
/..../\
/_./__/_\
A 4 B
|
|
D
```
Step 4: Let's use trigonometry to find the length of BN. In right triangle ABD, we have angle ADB = 90°. Since triangle ABC is isosceles, BAC = BCA = 30°. Thus, angle BAD = angle BDA = 75°.
Step 5: We can use the sine rule to find BN:
```
sin(BAD) / BD = sin(ABD) / AB
sin(75°) / 4 = sin(BDA) / 4
sin(75°) = sin(BDA)
```
Step 6: Since sin(75°) = cos(15°), we have:
```
cos(15°) = sin(BDA)
```
Step 7: Now, we can find the length of BN by using the cosine rule in triangle BDN:
```
BN^2 = BD^2 + DN^2 - 2 * BD * DN * cos(BDN)
Since BD = 4 and DN = 2 (since point N is the midpoint of CD),
BN^2 = 4^2 + 2^2 - 2 * 4 * 2 * cos(BDN)
BN^2 = 16 + 4 - 16 * cos(BDN)
BN^2 = 20 - 16 * cos(BDN)
```
Step 8: To calculate BN, we need to find the value of cos(BDN). We know that BDN = 90° - BAD.
```
cos(BDN) = cos(90° - BAD)
cos(BDN) = cos(90° - 75°)
cos(BDN) = cos(15°)
```
Step 9: Substitute cos(15°) back into the equation for BN:
```
BN^2 = 20 - 16 * cos(15°)
BN^2 = 20 - 16 * cos(15°)
```
Step 10: Calculating BN:
```
BN^2 = 20 - 16 * cos(15°)
BN^2 = 20 - 16 * 0.9659
BN^2 = 20 - 15.4558
BN^2 = 4.5442
BN = sqrt(4.5442)
BN ≈ 2.13
```
Thus, BN is approximately equal to 2.13 units.
To find BN, we need to consider the given information about the isosceles triangle ACB, with AB = AC = 4 and BAC = 30°. Let's visualize this triangle and its properties.
First, draw a straight line segment AC with length 4 units. Then, mark point B such that AB = AC = 4 units. This ensures that we have an isosceles triangle.
Next, construct a perpendicular line segment CD from the vertex C to the base AB. This is the altitude of the triangle.
Now, we need to find BN. To do so, we'll use some properties of the isosceles triangle.
In an isosceles triangle, the altitude will bisect the base. This means that CD divides AB into two equal parts. Let's mark the point of intersection between CD and AB as N. So, we can say that AN = NB.
To find the value of BN, we need to determine the length of AN. Since CD is the altitude, it forms a right angle with AB. Also, since BAC = 30°, the remaining angles in triangle ACB are each 75° (since the sum of angles in a triangle is 180°).
Now, we can apply trigonometric ratios to find AN. We have two angles - 75° and 30°, and we need to find the length of the opposite side AN.
Using the trigonometric ratio for the sine function, we can write:
sin(30°) = AN / AC
sin(30°) = AN / 4
We know that sin(30°) = 1/2, so we can substitute this value:
1/2 = AN / 4
Now, we can solve for AN:
AN = 1/2 * 4
AN = 2 units
Since AN = NB, BN is also equal to 2 units.
So, the length of BN in the given isosceles triangle is 2 units.