Question

the figure below is a right triangle with point q along ji. what would be the value of cos(∠ihq) - sin(∠qhj)?

Explanation:

Step1: Find the length of \(HJ\)

Using the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\), where \(c = 15\) (hypotenuse) and \(b=12\). Let \(a = HJ\). Then \(HJ=\sqrt{15^{2}-12^{2}}=\sqrt{(15 + 12)(15 - 12)}=\sqrt{27\times3}=\sqrt{81}=9\)

Step2: Define \(\cos(\angle IHJ)\) and \(\sin(\angle QHJ)\)

In right - triangle \(HJI\), \(\cos(\angle IHJ)=\frac{HJ}{JI}\) (adjacent over hypotenuse). Since \(HJ = 9\) and \(JI=15\), \(\cos(\angle IHJ)=\frac{9}{15}\)
\(\angle QHJ+\angle JQH = 90^{\circ}\) and \(\angle QHJ+\angle IHJ = 90^{\circ}\) (because \(\angle JHI = 90^{\circ}\)), so \(\angle QHJ\) and \(\angle IHJ\) are complementary. In a right - triangle, if \(\alpha+\beta = 90^{\circ}\), then \(\sin(\alpha)=\cos(\beta)\). So \(\sin(\angle QHJ)=\cos(\angle IHJ)\)

Step3: Calculate \(\cos(\angle IHJ)-\sin(\angle QHJ)\)

Substitute the values: \(\cos(\angle IHJ)-\sin(\angle QHJ)=\frac{9}{15}-\frac{9}{15}\)

Answer:

\(0\)