A kilogram %30 chocolate,
b kilogram %50 chocolate,
a+b = 100 kilogram chocolate,
48.(a+b) = 30.a + 50.b
48a + 48b = 30a + 50b
18a = 2b
b = 9a
a+b = a + 9a = 10a = 100 kg
a = 10 kg
b = 9.a = 90 kg
You should use,
10 kg %30 chocolate,
90 kg %50 chocolate
Given:
Angle A = 18.6°
Angle B = 93°
Length of side AB = 646 meters
To find:
the distance across the river, distance between BC
Steps:
Since we know the measure of 2 angles of a triangle we can find the measure of the third angle.
18.6° + 93° + ∠C = 180°
111.6° + ∠C = 180°
∠C = 180° - 111.6°
∠C = 68.4°
Therefore the measure of angle C is 68.4°.
now we can use the law of Sines,


![BC[sin(68.4)] = 646 [sin(18.6)]](https://tex.z-dn.net/?f=BC%5Bsin%2868.4%29%5D%20%3D%20646%20%5Bsin%2818.6%29%5D)



meters
Therefore, the distance across the river is 222 meters.
Happy to help :)
If anyone need more help, feel free to ask
Answer is a step by step so you divide