Constuct a triangle ABC ,given that angleA is 70 degree & angle C is 35 degree and length of perpendicular from A on BC is 5cm .meausre BC
Since A+B+C=180, angle B is 75°
BC = 5/sin75°
As for the construction,
draw a horizontal line AP
Construct angle A, labeling Q the end of the other ray than AP.
Construct a line XY parallel to AP, 5 units above it.
If necessary, extend AQ so it intersects XY at C.
Now construct angle C.
Point B is the intersection of that new ray with AP.
Sin A over a equels to sin B over b
Sin 70 over 5 equals to sin 75 over b
We cross multiply
bsin70 =5sin75
We divide by sin70
b=5sin75
sin70
b=5,14cm
I am not under stand
To construct triangle ABC, follow these steps:
1. Draw a line segment BC. This will be the base of the triangle.
2. At point B, draw a ray BA such that angle BAC measures 70 degrees.
3. At point C, draw a ray CA such that angle ACB measures 35 degrees.
4. Now, we need to find the length of BC given that the perpendicular from A on BC is 5 cm.
To find the length of BC, we can use the trigonometric relation involving the length of the perpendicular and the angles.
Let's define the length of BC as x.
- In triangle ABC, the perpendicular from A on BC divides BC into two segments. Let's call these segments BD and DC.
- Segment BD is the distance from B to the foot of the perpendicular, and segment DC is the distance from the foot of the perpendicular to C.
Now, using trigonometry, we can set up the following relation:
tan 70° = BD / 5 cm
The tangent of angle BAC is equal to the ratio of the length of BD to the length of the perpendicular.
Solving this equation for BD, we get:
BD = 5 cm * tan 70°
Now, we need to find the length of DC. Since the angles of a triangle add up to 180 degrees, we can find angle BCA by subtracting angles BAC and ACB from 180 degrees:
angle BCA = 180° - 70° - 35°
Now, we can use the sine rule to find DC:
sin angle BCA = DC / x
Solving this equation for DC, we get:
DC = x * sin angle BCA
Finally, the length of BC can be found by adding BD and DC:
BC = BD + DC
Substituting the values we have, we can find the length of BC.