Question

which of the following is closest to the value of x in th figure below? 10 of 11 questio 51.3 38.7 38.0 32.0

Explanation:

Step1: Recall the cosine formula

In a right - triangle, \(\cos x=\frac{\text{adjacent}}{\text{hypotenuse}}\). But wait, this is not a right - triangle. Let's use the Law of Cosines. For a triangle with sides \(a = 10\), \(b\), \(c = 16\) and angle \(x\). Wait, no, another approach. Let's drop a perpendicular from the right - angled vertex to the base of length \(16\). But a better way: assume it's a right - triangle (wait, no, the vertex is right - angled. Wait, no, the given triangle has a right - angle at the top vertex. Let's use the cosine formula. Wait, no, for a right - triangle (since there is a right - angle in the triangle), \(\cos x=\frac{\text{adjacent}}{\text{hypotenuse}}\) is wrong. Wait, in a right - triangle, \(\cos x=\frac{\text{adjacent}}{\text{hypotenuse}}\) is for non - right angles. Wait, the side adjacent to \(x\) is not \(10\). Wait, use the Law of Cosines: \(c^{2}=a^{2}+b^{2}-2ab\cos C\). But since there is a right - angle (\(90^{\circ}\)), we can use the area formula or trigonometric relations. Wait, no, let's use the Law of Cosines. Let the sides be \(a = 10\), \(b\), \(c = 16\). Wait, no, in a triangle with a right - angle (\(90^{\circ}\)), we can use the formula for \(\cos x\). Wait, the two legs of the right - triangle (the ones forming the right - angle) can be found using the area. Let the two legs be \(m\) and \(n\). The area \(S=\frac{1}{2}mn\). Also, using the formula \(S = \frac{1}{2}\times16\times h\) (where \(h\) is the altitude to the base of length \(16\)). But another approach: use the Law of Cosines. Let the sides be \(a = 10\), \(b\), \(c = 16\). Wait, no, in a triangle with angles \(x\), \(y\), \(90^{\circ}\) and sides opposite to them \(a\), \(b\), \(c\) (where \(c\) is the hypotenuse). Wait, no, the side opposite to \(x\) is not \(10\). Wait, use the Law of Cosines: \(16^{2}=10^{2}+b^{2}-2\times10\times b\times\cos(90^{\circ})\). Since \(\cos(90^{\circ}) = 0\), we have \(b=\sqrt{16^{2}-10^{2}}=\sqrt{256 - 100}=\sqrt{156}\approx12.5\). Now, use the Law of Cosines for angle \(x\): \(\cos x=\frac{16^{2}+10^{2}-b^{2}}{2\times16\times10}\). Substitute \(b^{2}=156\), \(\cos x=\frac{256 + 100-156}{320}=\frac{200}{320}=0.625\). Then \(x=\cos^{-1}(0.625)\approx51.3^{\circ}\)

Answer:

51.3