additional math - trigonometry , probably the most challenging question for mid year exam 2019

While preparing students for the mid year exam 2019, I stumbled upon a very challenging question, I have never came across any question remotely similar in my years of teaching, the setter for the additional math paper must be either very creative mathematically or he/she must have a lot of a-math resources at his/her disposal.

It took me a while to figure out how to start, but one of my genius student named Oscar from Bukit Panjang High manage to to beat me to it

Step 1 (line 2): You draw two right angle triangles (see left side of whiteboard in photo). fill in the known sides from the values in line one and calculate the unknown side using pythagoras theorem.

Step 2 (line 3): Use the cosine additional formula \displaystyle \cos \left( {A-B} \right)=\cos A\cos B+\sin A\sin B to expand the expression.

Step 3 (line 3 to 4):Not many people are aware that \displaystyle \cos \,\times {{\cos }^{{-1}}}=1, . For example \displaystyle \cos \,\times {{\cos }^{{-1}}}\frac{4}{{\sqrt{{17}}}}=\frac{4}{{\sqrt{{17}}}} and \displaystyle \sin \,\times \ {{\sin }^{{-1}}}\frac{4}{5}=\frac{4}{5}

Step 4 (Last two lines): Simplify the surds by making them into a single fraction. Althogh the answer in the answer sheet states \displaystyle \frac{8}{{5\sqrt{{17}}}}, I think it is better to rationalize the denominator by multiplying both the numerator and denominator by square root 17 and leave the answer as \displaystyle \frac{{8\sqrt{{17}}}}{{85}}