Question 7:

A triangle $ABC$ is to be drawn with $AB = 10$, $BC = 7$, and the angle at $A = \theta$. Of the two possible triangles, the larger has three times the area of the smaller one. Find the value of $\cos \theta$.

Solution:

By constructing the perpendicular and letting the base segments be $x$ and $2x$, we analyze the smaller right triangles. From the Pythagoras theorem on the two right triangles: $$\sqrt{49 - x^{2}} = \sqrt{100 - 4x^{2}}$$

Squaring and simplifying: $$49 - x^{2} = 100 - 4x^{2}$$ $$3x^{2} = 51 \quad \implies \quad x = \sqrt{17}.$$

Finally, we compute: $$\cos \theta = \frac{\sqrt{17}}{5}.$$

Answer: (d) $\tfrac{\sqrt{17}}{5}$