Solve a Triangle Given Its Perimeter, Altitude, and Angle

• The perimeter gives the equation: \[ a + b + c = p \tag{1} \]
• The area of the triangle using sides \(b\) and \(c\) is: \[ \text{Area} = \frac{1}{2} b c \sin A \]
• The area using base \(a\) and altitude \(h\) is: \[ \text{Area} = \frac{1}{2} h a \]
• Equating the two expressions for the area: \[ b c \sin A = h a \tag{2} \]
• Using the cosine law: \[ a^2 = b^2 + c^2 - 2 b c \cos A \tag{3} \]
• From equation (1): \(a = p - (b + c)\). Substitute into equation (3): \[ (p - (b + c))^2 = b^2 + c^2 - 2 b c \cos A \] After simplification: \[ p^2 + 2 b c - 2 p (b + c) = -2 b c \cos A \tag{4} \]
• Substitute \(a = p - (b + c)\) into equation (2): \[ b c \sin A = h p - h (b + c) \tag{5} \]
• Let \(Z = b + c\) and \(Y = b c\). Equations (4) and (5) become: \[ p^2 + 2Y - 2pZ = -2Y \cos A \] \[ Y \sin A = hp - hZ \]
• Substitute \(p = 100\), \(h = 18\), \(A = 56^\circ\): \[ -200Z + (2 + 2\cos 56^\circ)Y = -10000 \] \[ 18Z + Y \sin 56^\circ = 1800 \]
• Solving gives: \[ Z = 62.6456, \quad Y = 811.035 \] So \(b + c = 62.6456\) and \(b c = 811.035\).
• Solving the quadratic equation \(b^2 - 62.6456 b + 811.035 = 0\) gives: \[ b = 44.3643, \quad c = 18.2812 \]
• Find side \(a\) using the cosine law: \[ a = \sqrt{b^2 + c^2 - 2bc\cos 56^\circ} = 37.3543 \text{ units} \]

Using the Pythagorean theorem:

\[ h^2 = 30^2 + 40^2 \implies h = 50 \text{ cm} \]

The perimeter is:

\[ P = 30 + 40 + 50 = 120 \text{ cm} \]

Using the cosine law to find the base \(x\):

\[ x^2 = 10^2 + 10^2 - 2(10)(10)\cos 30^\circ \implies x \approx 5.18 \text{ m} \]

The perimeter is:

\[ P = 10 + 10 + 5.18 = 25.18 \text{ m} \]