Solve a Triangle Given Its Perimeter, Altitude, and Angle
In this tutorial, we solve triangles by finding all sides when the
perimeter, an altitude, and an angle are known.
Problem 1: Solve a Triangle Given Its Perimeter
Triangle \(ABC\) has a perimeter of \(100\) units. The altitude corresponding
to side \(a\) has length \(h = 18\) units, and the angle at vertex \(A\) is
\(56^\circ\). Find the lengths of all sides.
Solution
-
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
\[
Z = b + c, \quad Y = b c
\]
-
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:
\[
b + c = 62.6456, \quad b c = 811.035
\]
-
Solving the quadratic equation:
\[
b^2 - 62.6456 b + 811.035 = 0
\]
gives:
\[
b = 44.3643 \quad \text{or} \quad b = 18.2812
\]
-
Thus:
\[
b = 44.3643,\quad c = 18.2812
\]
-
Find side \(a\) using the cosine law:
\[
a = \sqrt{b^2 + c^2 - 2bc\cos 56^\circ}
\]
\[
a = 37.3543 \text{ units}
\]
Problem 2: Find the Perimeter of a Right Triangle
Find the perimeter of a right triangle with legs of lengths \(30\) cm and
\(40\) cm.
Solution
Using the Pythagorean theorem:
\[
h^2 = 30^2 + 40^2
\]
\[
h = 50 \text{ cm}
\]
The perimeter is:
\[
P = 30 + 40 + 50 = 120 \text{ cm}
\]
Problem 3: Find the Perimeter of an Isosceles Triangle
An isosceles triangle has two equal sides of \(10\) m and an included angle of
\(30^\circ\). Find its perimeter.
Solution
Using the cosine law to find the base \(x\):
\[
x^2 = 10^2 + 10^2 - 2(10)(10)\cos 30^\circ
\]
\[
x \approx 5.18 \text{ m}
\]
The perimeter is:
\[
P = 10 + 10 + 5.18 = 25.18 \text{ m}
\]
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