Year 10

The Peak of The Calvary Hill

Let's get creative. Let's suppose that I'm at constant speed of 1 unit per minute. So that would mean that the amount of distance it took for me to get somewhere is equal in numerical value to the number of minutes that it takes for me to get somewhere. Let's suppose these values:

From the Bus to the Husgado Cave - 20 minutes
From Husgado Cave to the Peak of Calvary Hill - 35 minutes
From the Peak of Calvary Hill to Jordan River - 50 minutes
From the Jordan River to the Bus - 25 minutes

This forms a quadrilateral, which may turn into a triangle if we consider the bus to the Husgado Cave to the Peak of Calvary Hill as one direct measurement. Meaning:

From the Bus to the Husgado Cave to the Peak of Calvary Hill - 55 minutes

To be able to use the laws of sine and cosine, let's assign them as sides a a, b, and c, with us needing to look for angles A, B, and C.

a = 55 | b = 50 | c = 25

(a^2 - b ^2 - c^2)/-2bc = cosA
(65^2 - 50 ^2 - 20 ^2)/-2*50*20 = cosA
-100/-2000 = cosA
A = 87 degrees

55/sin87 = 50/sinB
(sin87*50)/55 = sinB
0.9 = sinB
B = 64 degrees

C = 180 - A - B
C = 180 - 64 - 87
C = 29 degrees

FINAL SOLUTION SET

A = 87 degrees | B = 64 degrees | C = 29 degrees
a = 55 | b = 50 | c = 25