Question

find m∠b.
c to b is a side, c to a is 25√2, a to b is 25√6, angle at a is a right angle.
m∠b = °

Explanation:

Step1: Identify triangle type

Triangle \( ABC \) is right - angled at \( A \), so we can use trigonometric ratios. Let's recall that in a right - triangle, \(\sin B=\frac{\text{opposite}}{\text{hypotenuse}}\) and \(\cos B = \frac{\text{adjacent}}{\text{hypotenuse}}\), \(\tan B=\frac{\text{opposite}}{\text{adjacent}}\). Here, the side opposite to \(\angle B\) is \( AC = 25\sqrt{2}\) and the side adjacent to \(\angle B\) is \( AB=25\sqrt{6}\). We can also use the tangent function: \(\tan B=\frac{AC}{AB}\)

Step2: Calculate \(\tan B\)

Substitute the values of \( AC \) and \( AB \) into the tangent formula:
\(\tan B=\frac{25\sqrt{2}}{25\sqrt{6}}\)
Simplify the fraction by canceling out the common factor \( 25 \):
\(\tan B=\frac{\sqrt{2}}{\sqrt{6}}\)
Rationalize the denominator by multiplying the numerator and denominator by \(\sqrt{6}\):
\(\tan B=\frac{\sqrt{2}\times\sqrt{6}}{\sqrt{6}\times\sqrt{6}}=\frac{\sqrt{12}}{6}=\frac{2\sqrt{3}}{6}=\frac{\sqrt{3}}{3}\)

Step3: Find the measure of \(\angle B\)

We know that if \(\tan B = \frac{\sqrt{3}}{3}\), then we need to find the angle whose tangent is \(\frac{\sqrt{3}}{3}\). From the special - angle trigonometric values, we know that \(\tan30^{\circ}=\frac{\sqrt{3}}{3}\). So, \(m\angle B = 30^{\circ}\)

Answer:

\(30\)