Answer:
(a) 315°
(b) 3°
(c) 238°
Step-by-step explanation:
Bearings are measured clockwise from north. The triangle described is illustrated in the attachment.
<h3>(a)</h3>
The bearing of P from R is 180° different from the bearing of R from P it will be ...
135° +180° = 315° . . . . bearing of P from R
__
<h3>(b)</h3>
The bearing of Q from R is 48° more than the bearing of P from R, so is ...
315° +48° = 363°, or 3° . . . . bearing of Q from R
__
<h3>(c)</h3>
The angle QPR has a value that makes the sum of angles in the triangle equal to 180°. It is ...
180° -48° -55° = 77°
The bearing of Q from P is 77° less than the bearing of R from P, so is ...
135° -77° = 58°
As above, the reverse bearing from Q to P is ...
58° +180° = 238° . . . . bearing of P from Q
Answer:
∠x = 90°
∠y = 58°
∠z = 32°
Step-by-step explanation:
he dimensions of the angles given are;
∠B = 32°
Whereby ΔABC is a right-angled triangle, and the square fits at angle A, we have;
∠A = 90°
∠B + ∠C = 90° which gives
32° + ∠C = 90°
∠C = 58°
∠x + Interior angle of the square = 180° (Sum of angles on a straight line)
∠x + 90° = 180°
∠x = 90°
∠x + ∠y + 32° = 180° (Sum of angles in a triangle)
90° + ∠y + 32° = 180°
∠y = 180 - 90° - 32° = 58°
∠y + ∠z + Interior angle of the square = 180° (Sum of angles on a straight line)
58° + ∠z +90° = 180°
∴ ∠z = 32°
∠x = 90°
∠y = 58°
∠z = 32°
Answer:
31
Step-by-step explanation:
Given
x² + 5x - 3 = 0
with a = 1, b = 5, c= - 3 , then
sum of roots x₁ + x₂ = -
= -
= - 5
product of roots =
=
= - 3
Now
(x₁ + x₂)² = x₁² + 2x₁x₂ + x₂² , that is
(- 5)² = x₁² + 2(- 3) + x₂²
25 = x₁² - 6 + x₂² ( add 6 to both sides ), then
x₁² + x₂² = 31
Let
The origin of coordinates the tree
r1 = vector position of the child 1.
r2 = vector position of the child 2
Child 1:
r1 = (12i + 12j)
Child 2:
r2 = (-18i + 11j)
The scalar product will be given by:
r1.r2 = ((12) * (- 18)) + ((12) * (11)) = - 84
The scalar product of their net displacements from the tree is -84m ^ 2