The party of kejriwal walks from gwalior to Bhuvneshwar. They first walk straight for 150km and then turn 45 ddegree towards Varansi and again go straight for 120km how for have they travelled from the initial place. Please show the method.

say tthey startedd east

east distance = 150 + 120 cos 45
north distance = 120 sin 45

hypotenuse =
sqrt[(150+120cos45)^2+(120sin 45)^2]

missing the distance along the 45° path

I read the question as consisting of 3 legs of the journey.

Funny thing is that I answered the same question back in February, as seen from the Related Question below.
http://www.jiskha.com/display.cgi?id=1486127947

Damon's answer yields the same result

To determine how far the party of Kejriwal has traveled from the initial place, we need to use certain geometric principles.

Let's visualize the situation. The party first walks straight for 150km, which we can represent as a line segment. Then, they turn 45 degrees towards Varanasi, forming a right triangle.

Using trigonometry, we can find the distance the party has traveled since the initial place.

Step 1: Calculate the distance covered after turning towards Varanasi.
- The distance covered in the direction towards Varanasi is the adjacent side of the right triangle formed.
- We can find this distance using the formula: adjacent side = 120 km * cos(45°).

Evaluating this calculation: adjacent side = 120 km * cos(45°) ≈ 84.85 km.

Step 2: Calculate the total distance from the initial place.
- The total distance is the sum of the straight-line distance of 150 km and the adjacent side distance of 84.85 km.

Evaluating this calculation: total distance = 150 km + 84.85 km ≈ 234.85 km.

Therefore, the party of Kejriwal has traveled approximately 234.85 km from the initial place.