From ② and ①, the sin θ cancels out.

B ÷ A :

y/x = r sinθsinΦ ÷ r sinθcosΦ

y/x = sinΦ/cosΦ = tanΦ

Φ = arctan(y/x) or tan-¹(y/x)



A² + B² :

x² + y² = r² sin²θcos²Φ + r² sin²θsin²Φ

x² + y² = r²sin²θ(cos²Φ + sin²Φ)

x² + y² = r²sin²θ

(recall trigo property cos²x + sin²x = 1. So recognising this from the start, we square the terms and factor to eliminate one variable Φ)


Now, sub C in by squaring first.

C² :

z² = r²cos²θ

so from above result,

A² + B² + C² :

x² + y² + z² = r²sin²θ + r²cos²θ = r²

r = √(x² + y² + z²)



Since we know r already, sub back into C.

z = √(x² + y² + z²)sinθ

sinθ = z/√(x² + y² + z²)

(part b is solved first)



Then, θ = arcsin(z/√(x² + y² + z²))

Or sin-¹(z/√(x² + y² + z²))