Question

a. do not solve the following triangles. simply list a formula/law that will solve for each of the these unknowns in each of the following triangles. fill in the givens in each of your formula. use the most efficient method. (3 marks each)
d

Explanation:

Left Triangle (Triangle DEF):

Step1: Law of Sines

The Law of Sines is given by \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\). In triangle \(DEF\), if we want to find side \(d\) (opposite the angle we can find using the angle - sum property of a triangle) or other unknowns, we can use \(\frac{11}{\sin F}=\frac{d}{\sin70^{\circ}}=\frac{18}{\sin D}\) (where \(D + 70^{\circ}+F=180^{\circ}\)).

Right Triangle (Triangle PQR):

Step1: Pythagorean Theorem

For the side \(q\) (the hypotenuse), we use the Pythagorean Theorem \(a^{2}+b^{2}=c^{2}\). Here, \(q=\sqrt{8^{2}+15^{2}}\).

Step2: Trigonometric Ratios (for the non - right angles)

For the angle at \(R\), we can use \(\sin R=\frac{8}{q}\) (after finding \(q\) from the Pythagorean Theorem) or \(\cos R=\frac{15}{q}\), and for the angle at \(P\), \(\sin P=\frac{15}{q}\) or \(\cos P=\frac{8}{q}\).

Answer:

Left Triangle: \(\frac{11}{\sin F}=\frac{d}{\sin70^{\circ}}=\frac{18}{\sin D}\) (with \(D + 70^{\circ}+F = 180^{\circ}\)); Right Triangle: \(q=\sqrt{8^{2}+15^{2}}\), \(\sin R=\frac{8}{q}\), \(\cos R=\frac{15}{q}\), \(\sin P=\frac{15}{q}\), \(\cos P=\frac{8}{q}\) (where \(q\) is found from the Pythagorean Theorem first).