Question

solve the given right triangle for its missing angle and side measures.
note: figure not drawn to scale.
a.
m∠d = 35°, de = 8.40 units, df = 13.65 units
b.
m∠d = 55°, de = 8.40 units, df = 14.65 units
c.
m∠d = 35°, de = 8.40 units, df = 14.65 units
d.
m∠d = 55°, de = 4.40 units, df = 13.65 units

Explanation:

Step1: Find angle D

In a right triangle, the sum of angles is \(180^\circ\). Since \(\angle E = 90^\circ\) and \(\angle F = 35^\circ\), we have \(m\angle D=180^\circ - 90^\circ - 35^\circ = 55^\circ\).

Step2: Find DE (opposite to \(\angle F\))

Using sine: \(\sin(35^\circ)=\frac{DE}{EF}\)? Wait, no, \(EF = 12\) (adjacent to \(\angle F\)), \(DE\) is opposite to \(\angle F\). Wait, \(\tan(35^\circ)=\frac{DE}{EF}\)? No, \(EF = 12\) (adjacent to \(\angle F\)), \(DE\) is opposite, \(DF\) is hypotenuse. Wait, \(\sin(35^\circ)=\frac{DE}{DF}\), \(\cos(35^\circ)=\frac{EF}{DF}\), \(\tan(35^\circ)=\frac{DE}{EF}\). Since \(EF = 12\) (adjacent to \(\angle F\)), \(DE=\tan(35^\circ)\times EF\). \(\tan(35^\circ)\approx0.7002\), so \(DE\approx0.7002\times12\approx8.40\).

Step3: Find DF (hypotenuse)

Using cosine: \(\cos(35^\circ)=\frac{EF}{DF}\), so \(DF=\frac{EF}{\cos(35^\circ)}\). \(\cos(35^\circ)\approx0.8192\), \(EF = 12\), so \(DF\approx\frac{12}{0.8192}\approx14.65\).

Answer:

B. \(m\angle D = 55^\circ\), \(DE\approx8.40\) units, \(DF = 14.65\) units