Question

select the correct answer from each drop - down menu.
in \\( \triangle a b c, m \angle b = 90 ^ { \circ }, \cos ( c ) = \frac { 1 5 } { 1 7 } \\), and \\( b c = 3 0 \\) units.
based on this information, \\( m \angle a = \square ^ { \circ }, m \angle c = \square ^ { \circ } \\), and \\( a b = \square \\) units.
note that the angle measures are rounded to the near
hint: first solve for \\( m \angle c \\) by using the cosine equatio
33 62 28 degree
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Explanation:

Step1: Find \(m\angle C\)

Given \(\cos(C)=\frac{15}{17}\). Using the inverse - cosine function \(C = \cos^{-1}(\frac{15}{17})\).

$$C=\cos^{-1}(\frac{15}{17})\approx28^{\circ}$$

Step2: Find \(m\angle A\)

In \(\triangle ABC\), since \(m\angle B = 90^{\circ}\), and the sum of angles in a triangle is \(180^{\circ}\). So \(m\angle A=180^{\circ}-m\angle B - m\angle C\).

$$m\angle A = 180^{\circ}-90^{\circ}-28^{\circ}=62^{\circ}$$

Step3: Find \(AB\)

Using the tangent function \(\tan(C)=\frac{AB}{BC}\). Since \(BC = 30\) and \(\tan(28^{\circ})\approx0.5317\) (from trigonometric tables or calculator), and \(\tan(C)=\frac{AB}{BC}\), then \(AB = BC\times\tan(C)\).

$$AB = 30\times\tan(28^{\circ})\approx30\times0.5317 = 16$$

Answer:

\(m\angle A = 62^{\circ}\), \(m\angle C=28^{\circ}\), \(AB = 16\) units.