Question

20)
(there is a right triangle abc with right angle at c, angle at b is 53 degrees, and side ab (the hypotenuse) is 5 m long. the problem is likely about finding the lengths of other sides or other angle, but the specific question is not provided in the ocr text.)

Explanation:

Since the problem is about a right - angled triangle (angle \(C = 90^{\circ}\), \(AB = 5\space m\), \(\angle B=53^{\circ}\)) and we might want to find the lengths of other sides or other angles, we can use trigonometric relationships. Let's assume we want to find the length of \(BC\) (adjacent side to \(\angle B\)) or \(AC\) (opposite side to \(\angle B\)) or angle \(A\).

Step 1: Find angle \(A\)

In a triangle, the sum of interior angles is \(180^{\circ}\). For right - angled triangle \(ABC\) with \(\angle C = 90^{\circ}\) and \(\angle B=53^{\circ}\), we have:
\(\angle A+\angle B+\angle C = 180^{\circ}\)
Substitute \(\angle B = 53^{\circ}\) and \(\angle C=90^{\circ}\) into the formula:
\(\angle A=180^{\circ}-\angle B - \angle C=180^{\circ}- 53^{\circ}-90^{\circ}=37^{\circ}\)

Step 2: Find the length of \(BC\) (adjacent side to \(\angle B\))

We know that \(\cos B=\frac{BC}{AB}\), where \(AB = 5\space m\) and \(\angle B = 53^{\circ}\)
So, \(BC = AB\times\cos B\)
We know that \(\cos53^{\circ}\approx0.6\) (approximate value, since \(\cos53^{\circ}\approx\frac{3}{5}\)) and \(AB = 5\space m\)
\(BC=5\times\cos53^{\circ}\approx5\times0.6 = 3\space m\)

Step 3: Find the length of \(AC\) (opposite side to \(\angle B\))

We know that \(\sin B=\frac{AC}{AB}\)
So, \(AC = AB\times\sin B\)
We know that \(\sin53^{\circ}\approx0.8\) (approximate value, since \(\sin53^{\circ}\approx\frac{4}{5}\)) and \(AB = 5\space m\)
\(AC = 5\times\sin53^{\circ}\approx5\times0.8=4\space m\)

If we assume the question is to find the length of \(BC\), the answer is \(3\space m\); if to find \(AC\), the answer is \(4\space m\); if to find \(\angle A\), the answer is \(37^{\circ}\)

Answer:

Since the problem is about a right - angled triangle (angle \(C = 90^{\circ}\), \(AB = 5\space m\), \(\angle B=53^{\circ}\)) and we might want to find the lengths of other sides or other angles, we can use trigonometric relationships. Let's assume we want to find the length of \(BC\) (adjacent side to \(\angle B\)) or \(AC\) (opposite side to \(\angle B\)) or angle \(A\).

Step 1: Find angle \(A\)

In a triangle, the sum of interior angles is \(180^{\circ}\). For right - angled triangle \(ABC\) with \(\angle C = 90^{\circ}\) and \(\angle B=53^{\circ}\), we have:
\(\angle A+\angle B+\angle C = 180^{\circ}\)
Substitute \(\angle B = 53^{\circ}\) and \(\angle C=90^{\circ}\) into the formula:
\(\angle A=180^{\circ}-\angle B - \angle C=180^{\circ}- 53^{\circ}-90^{\circ}=37^{\circ}\)

Step 2: Find the length of \(BC\) (adjacent side to \(\angle B\))

We know that \(\cos B=\frac{BC}{AB}\), where \(AB = 5\space m\) and \(\angle B = 53^{\circ}\)
So, \(BC = AB\times\cos B\)
We know that \(\cos53^{\circ}\approx0.6\) (approximate value, since \(\cos53^{\circ}\approx\frac{3}{5}\)) and \(AB = 5\space m\)
\(BC=5\times\cos53^{\circ}\approx5\times0.6 = 3\space m\)

Step 3: Find the length of \(AC\) (opposite side to \(\angle B\))

We know that \(\sin B=\frac{AC}{AB}\)
So, \(AC = AB\times\sin B\)
We know that \(\sin53^{\circ}\approx0.8\) (approximate value, since \(\sin53^{\circ}\approx\frac{4}{5}\)) and \(AB = 5\space m\)
\(AC = 5\times\sin53^{\circ}\approx5\times0.8=4\space m\)

If we assume the question is to find the length of \(BC\), the answer is \(3\space m\); if to find \(AC\), the answer is \(4\space m\); if to find \(\angle A\), the answer is \(37^{\circ}\)