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ab = m∠a = m∠b = system of equations save copy 1 y= 2 y= -10 0

Question

ab =

m∠a =

m∠b =

system of equations save copy

1

y=

2

y=

-10 0

Explanation:

Step1: Use Pythagorean theorem to find \(AB\)

In right - triangle \(ABC\) with \(AC = 4.2m\), \(BC=6.5m\) and \(\angle C = 90^{\circ}\). By Pythagorean theorem \(AB^{2}=AC^{2}+BC^{2}\).

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Step2: Use trigonometric ratio to find \(\angle A\)

We know that \(\tan A=\frac{BC}{AC}\). Since \(BC = 6.5m\) and \(AC = 4.2m\), \(\tan A=\frac{6.5}{4.2}\approx1.5476\). Then \(A=\arctan(1.5476)\approx57.1^{\circ}\)

Step3: Use the property of right - triangle angles to find \(\angle B\)

In right - triangle \(ABC\), \(\angle A+\angle B = 90^{\circ}\). Since \(\angle A\approx57.1^{\circ}\), then \(\angle B=90^{\circ}-\angle A\approx90 - 57.1=32.9^{\circ}\)

Answer:

\(AB\approx7.74m\), \(m\angle A\approx57.1^{\circ}\), \(m\angle B\approx32.9^{\circ}\)