Correct answer is : Area of triangle is ![15\sqrt{3} sq.units](https://tex.z-dn.net/?f=15%5Csqrt%7B3%7D%20sq.units)
Solution:-
We can do it in 2 methods.
Method 1:-
Given that AB=6, BC=10 and m∠B = 120
Then area of triangle = ![\frac{1}{2}XbaseXheight](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7DXbaseXheight)
Let us assume AB is base and D is an altitude from C onto AB.
Then sin(60)= ![\frac{CD}{BC}](https://tex.z-dn.net/?f=%5Cfrac%7BCD%7D%7BBC%7D)
CD = BC sin(60)
Hence height =
= ![5\sqrt{3}](https://tex.z-dn.net/?f=5%5Csqrt%7B3%7D)
Hence area of ΔABC =
sq.units
Method2:-
Area of triangle = ![\frac{1}{2} acsin(B)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20acsin%28B%29)
Here a= BC=10, c=AB=6.
Hence area of triangle =
sq.units
Given:
![m=0.56x-0.25](https://tex.z-dn.net/?f=m%3D0.56x-0.25)
![n=0.20x-0.03](https://tex.z-dn.net/?f=n%3D0.20x-0.03)
To find:
The sum of m and n.
Solution:
We have,
![m=0.56x-0.25](https://tex.z-dn.net/?f=m%3D0.56x-0.25)
![n=0.20x-0.03](https://tex.z-dn.net/?f=n%3D0.20x-0.03)
The sum of m and n is:
![m+n=(0.56x-0.25)+(0.20x-0.03)](https://tex.z-dn.net/?f=m%2Bn%3D%280.56x-0.25%29%2B%280.20x-0.03%29)
Combining the like terms, we get
![m+n=(0.56x+0.20x)+(-0.25-0.03)](https://tex.z-dn.net/?f=m%2Bn%3D%280.56x%2B0.20x%29%2B%28-0.25-0.03%29)
![m+n=(0.76x)+(-0.28)](https://tex.z-dn.net/?f=m%2Bn%3D%280.76x%29%2B%28-0.28%29)
![m+n=0.76x-0.28](https://tex.z-dn.net/?f=m%2Bn%3D0.76x-0.28)
Therefore, the sum of m and n is
.
Answer:
C
Step-by-step explanation:
I took the test
3 cm on the map represents 31.5 km in reality.
3cm : 31.5 km
1cm : 31.5/3 km
1cm : 10.5km
So the scale on the map is 1cm represents 10.5 km.