Question

6- in abc angle a=45 and angle b=30, the altitude of ab has a length of 10, find perimeter of δabc,

Explanation:

Step1: Define the triangle and altitude

Let's consider \(\triangle ABC\) with \(\angle A = 45^\circ\), \(\angle B=30^\circ\), and let the altitude from \(C\) to \(AB\) be \(CD = 10\) (where \(D\) is on \(AB\)). So we have two right triangles \(\triangle ACD\) and \(\triangle BCD\).

Step2: Find \(AC\) from \(\triangle ACD\)

In right triangle \(\triangle ACD\), \(\sin A=\frac{CD}{AC}\). Given \(\angle A = 45^\circ\) and \(CD = 10\). Since \(\sin45^\circ=\frac{\sqrt{2}}{2}\), we have \(\frac{\sqrt{2}}{2}=\frac{10}{AC}\). Solving for \(AC\), we get \(AC = \frac{10\times2}{\sqrt{2}}=10\sqrt{2}\). Also, \(\tan A=\frac{CD}{AD}\), and \(\tan45^\circ = 1=\frac{10}{AD}\), so \(AD = 10\).

Step3: Find \(BC\) from \(\triangle BCD\)

In right triangle \(\triangle BCD\), \(\sin B=\frac{CD}{BC}\). Given \(\angle B = 30^\circ\) and \(CD = 10\). Since \(\sin30^\circ=\frac{1}{2}\), we have \(\frac{1}{2}=\frac{10}{BC}\), so \(BC = 20\). Also, \(\tan B=\frac{CD}{BD}\), and \(\tan30^\circ=\frac{1}{\sqrt{3}}=\frac{10}{BD}\), so \(BD = 10\sqrt{3}\).

Step4: Find \(AB\)

\(AB=AD + BD=10 + 10\sqrt{3}\).

Step5: Find the perimeter

Perimeter \(P=AB + BC+AC=(10 + 10\sqrt{3})+20 + 10\sqrt{2}=30 + 10\sqrt{3}+10\sqrt{2}\).

Answer:

The perimeter of \(\triangle ABC\) is \(30 + 10\sqrt{2}+10\sqrt{3}\) (or approximately \(30+14.14 + 17.32=61.46\) if decimal approximation is needed).