If θ lies in the first quadrant and cos θ = 8/17,

According to the question,

cos θ = 8/17

sin θ = ±√(1 – cos2θ)

Since, θ lies in first quadrant, only positive sign can be considered.

⇒ sin θ = √(1 – 64/289) = 15/17

Let, y = cos(30° + θ) + cos (45° – θ) + cos (120° – θ)

We know that,

cos(x + y) = cos x cos y – sin x sin y

Therefore,

y = cos30° cos θ – sin30° sin θ + cos45° cos θ + sin45°sin θ +cos120° cos θ + sin120° sin θ

Substituting values of cos30°, sin30°, cos 120°, sin120° and cos 45°

$\Rightarrow y=\frac{\sqrt{3}}{2} \cdot \frac{8}{17}-\frac{1}{2} \cdot \frac{15}{17}+\frac{1}{\sqrt{2}} \cdot \frac{8}{17}+\frac{1}{\sqrt{2}} \cdot \frac{15}{17}+\left(-\frac{1}{2}\right)\left(\frac{8}{17}\right)+\left(\frac{\sqrt{3}}{2}\right)\left(\frac{15}{17}\right)$

$=\frac{8 \sqrt{3}}{34}-\frac{15}{34}+\frac{8+15}{17 \sqrt{2}}-\frac{8}{34}+\frac{15 \sqrt{3}}{34}$

$=\frac{23 \sqrt{3}}{34}+\frac{23}{17 \sqrt{2}}-\frac{23}{34}$

$\Rightarrow y=\frac{23}{34}(\sqrt{3}+\sqrt{2}-1)$