Question

Help! Thanks!!!!!!!!!

Answer 1

1)

∠BAC = ∠NAC - ∠NAB = 144 - 68 = 76⁰

AB = 370 m

AC = 510 m

To find BC we can use cosine law.

a² = b² + c² -2bc*cos A

|BC|² = |AC|²+|AB|² - 2|AC|*|AB|*cos(∠BAC)

|BC|² = 510²+370² - 2*510*370*cos(∠76⁰) =

|BC| ≈ 553 m

2)

To find ∠ACB, we are going to use law of sine.

sin(∠BAC)/|BC| = sin(∠ACB)/|AB|

sin(76⁰)/553 m = sin(∠ACB)/370 m

sin(∠ACB)=(370*sin(76⁰))/553 =0.6492

∠ACB = 40.48⁰≈ 40⁰

3)

∠BAC = 76⁰

∠ACB = 40⁰

∠CBA = 180-(76+40) = 64⁰

Bearing C from B =360⁰- 64⁰-(180-68) = 184⁰

4)

Shortest distance from A to BC is height (h) from A to BC.

We know that area of the triangle

A= (1/2)|AB|*|AC|* sin(∠BAC) =(1/2)*370*510*sin(76⁰).

Also, area the same triangle

A= (1/2)|BC|*h = (1/2)*553*h.

So, we can write

(1/2)*370*510*sin(76⁰) =(1/2)*553*h

370*510*sin(76⁰) =553*h

h= 370*510*sin(76⁰) / 553= 331 m

h=331 m