1. In a triangle, the measure of the first angle is three times the measure of the second angle. The measure of the third angle is 80° more than the measure of the second angle. Use the fact that the sum of the measures of the three angles of a triangle is 180° to find the measure of each angle.
2. A circle with radius r is inscribed into a right triangle. Find the perimeter of the triangle if:the length of the hypotenuse is 24 cm, and r=4 cm;
3. Right triangle ABC has legs AC and BC of lengths 16mi and 24mi accordingly. Find the distance of point B from the line that contains the median AM.
4. Given triangle ABC, AD is perpendicular to BC such that DC = 2 and BD = 3. If angle BAC = 45 degrees, then find the area of triangle ABC.
5. CPhill has answered all your others but I expect you can do the last one by yourself. - Melody.
+128475 1. In a triangle, the measure of the first angle is three times the measure of the second angle. The measure of the third angle is 80° more than the measure of the second angle. Use the fact that the sum of the measures of the three angles of a triangle is 180° to find the measure of each angle.
Let the second angle = A
The first angle = 3A
And the remaining angle = A + 80
So
A + 3A + A + 80 = 180 simplify
5A + 80 = 180 subtract 80 from both sides
5A = 100 divide both sides by 5
A = 20
Second angle = 60
Remaining angle = 100
+128475 Let C =(0,0)
Let A = (0,16)
Let B = (24,0)
The median, M = (12,0)
AM has the slope (16-0)/ ( 0-12) = -16/12= -4/3
And the equation of AM is
y =(-4/3)x + 16
In stadard form, Ax + By - C = 0, the equation is
3y = -4x + 48
4x + 3y - 48 = 0
The distance from B = (24,0) to this line is
l 4 (24) + 3(0) - 48 l l 48 l
___________________ = ---------- = 48/5 mi = 9.6 mi
sqrt ( 4^2 + 3^2 ] 5
+128475
4. Given triangle ABC, AD is perpendicular to BC such that DC = 2 and BD = 3. If angle BAC = 45 degrees, then find the area of triangle ABC.
We have something like this :
A
B 3 D 2 C
Using the Pythagorean Theorem :
AB^2 = BD^2 + AD^2
AC^2 = DC^2 + AD^2
AB^2 = 9 + AD^2
AC^2 = 4 + AD^2 subtract these
AB^2 - AC^2 = 5
AB^2 = 5 + AC^2
AB = sqrt (5 + AC^2)
Using the Law of Cosines
BC^2 = AB^2 + AC^2 - 2 (AB * AC)cos (45°)
Let x = AC
5^2 = (5 + x^2) + x^2 - 2* [ sqrt (5 + x^2) * x] (sqrt (2) /2 )
25 = 5 + x^2 + x^2 - sqrt (2) [ sqrt (5 + x^2) * x ]
Solving this for the positive value of x gives x = 2sqrt (10) = sqrt (40) = AC
And AB = sqrt (5 + AC^2) = sqrt (5 + 40) = sqrt (45)
So....the area of triangle ABC =
(1/2) AC * AB sin (45°) =
(1/2) (sqrt (40) * sqrt (45) * [ sqrt (2) / 2 ] =
(1/4) * sqrt (3600) =
(1/4) * 60 =
15 units^2
+128475
2. A circle with radius r is inscribed into a right triangle. Find the perimeter of the triangle if:the length of the hypotenuse is 24 cm, and r=4 cm;
See the following image :
Let
AE = 4
EC = a
CG = a
BG = b
BF = b
FA = 4
So the sides are
a + b = 24
b + 4
4 + a
And
a + b = 24 → b = 24 - a
b + 4 → 28 - a
4 + a
Using the Pythagorean Theorem :
(4 + a)^2 + ( 28 - a)^2 = 24^2
Solving this for a produces
a = 4 ( 3 + sqrt (2) ) = 12 + 4sqrt (2)
4 + a = 16 + 4sqt (2)
b = 24 - a = 24 - (12 + 4 sqrt (2) ) = 12 - 4sqrt (2)
And b + 4 = 16 - 4sqrt (2)
So....the perimeter =
24 + (16 - 4sqrt (2)) + (16 + 4sqrt (2) ) =
24 + 32
56 units
