Question

Someone please solve and explain part a(i)

Answer 1

Step-by-step explanation:

the way I understand the description :

C is below B. they are on a kind of straight hill, and there is a straight "road" going up from C to B.

then, at B there is an antenna or other firm of mast going straight up.

therefore, this is not a right-angled triangle with 90 degrees at B (as it would be, if this would be in a flat plane).

but because it goes downhill from B to C the angle is 105 degree.

(a)(i)

now, imagine, there would be a horizontal plane either at B or at C. AB would have a true 90 degree angle with this plane. so, what is the angle of CB with this plane ?

this angle is the "excess" of the 90 degrees, as CB angles down from the horizontal plane at B, or angles up with the same angle from the horizontal plane at C.

what is the "excess" of 105 degrees vs. the standard 90 degrees ? 105 - 90 = 15 degrees.

(a)(ii)

the extended Pythagoras for not right-angled triangles :

c² = a² + b² - 2ab×cos(B)

B being the angle opposite of the Hypotenuse c.

so, we have

c² = 15² + 10² -2×15×10×cos(105) = 225 + 100 - 300×cos(105) =

= 325 - 300×cos(105) = 402.6457135...

c = 20.06603383... ≈ 20 m