Question

A person A and B travel from same point the person A moves 8 km to the south and turns at angle 30 degree . Person B moves about x km towards east and turns 60 degree and heads towards north . Then find out at what height two men meet from origin and distance travelled by person B before he turns to north

Answer 1

Answer:

Step-by-step explanation:

Since PERSON a moves 8 km

8 × sin60 =( x) × sin 60

Therefore x = 8 km

PERSON b also moves 8 km towards east

Now since a moves 30° and b moves 60° now remaining Angle is 90°

Therefore its consider has right angled triangle

A^2 +B^2 =C^2

8+8 = C

C = 16

Distance travelled by b is 16 km

Answer 2

Distance travelled by B before turn is 11.31 km

Step-by-step explanation:

Since Person A is moving 8 km to South (Let Denote that by a) $8 \times sin60$°

Which will be equal to that travelled  by B i.e. $x \times sin60$°

According to the Question -

$8 \times sin60 = x \times sin60$°

x = 8 km

Now the movement of A & B forms a Right Angled Triangle, with h being the Hypotenuse

• Applying Pythagoras Theorem, we get,

*Refer Figure Attached for Easy understanding,

$a^2 + b^2 = h^2$

⇒ $8^2 + 8^2 = h^2$

⇒ $h = \sqrt {8^2 + 8^2}$

⇒ $h = \sqrt {64+64}$

⇒ $h = \sqrt {128}$

⇒ h = 11.31 km

Thus, Distance travelled by B before turn will be 11.31 km.