Find, correct to the nearest degree, the three angles of the triangle with the given vertices. p(1, 0), q(0, 1), r(4, 4)

Mathematics

1 answer:

Vilka [71]3 years ago

3 0

A geometry app shows the angles to be
∠p ≈ 82°
∠q ≈ 82°
∠r ≈ 16°

_____
The length of pq is √2, and the length of pr is 5. Then
cos(p) = √2/(2·5)
p = arccos(√2/10) = 81.87°
The triangle is isosceles, so q = 81.87° and r = 180° -2*81.87° = 16.26°.

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Δ MAL and Δ DLA are right triangles. If Δ MAL ≅ ΔDLA, then side DL is congruent to side MA, and ∠ MAL is congruent to ∠ DLA. Giv

nevsk [136]

Answer:

See below.

Step-by-step explanation:

Given :

ΔMAL ≅ ΔDLA, DL = MA, ∠MAL = ∠DLA
∠M = 30°
DL = (2x + 10) cm
MA = (3x - 2) cm
AL = (x + 5) cm

To Find :

DL
AL
∠DLA
∠ADL

Solving :

DL = MA
2x + 10 = 3x - 2
x = 12
DL = 24 + 10 = 34 cm

AL = x + 5
AL = 12 + 5
AL = 17 cm

∠DLA = 180° - 90° - 30° = 180° - 120° = 60°

∠ADL = ∠M = 30°

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Answer:

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Solve, using the substitution method.

user100 [1]

$4x+2y=11 \\
x-2=-2y \\ \\
\hbox{from the 2nd equation:} \\
x-2=-2y \\
x=-2y+2 \\ \\
\hbox{substitute -2y+2 for x in the 1st equation:} \\
4(-2y+2)+2y=11 \\
-8y+8+2y=11 \\
-8y+2y=11-8 \\
-6y=3 \\
y=-\frac{3}{6}=-\frac{1}{2} \\ \\
\hbox{solve one of the equations for x:} \\
x=-2 \cdot (-\frac{1}{2}) +2=1+2=3 \\ \\ \hbox{the answer:} \\ x=3 \\ y=-\frac{1}{2}$