We have that
csc ∅=13/12
sec ∅=-13/5
cot ∅=-5/12
we know that
csc ∅=1/sin ∅
sin ∅=1/ csc ∅------> sin ∅=12/13
sec ∅=-13/5
sec ∅=1/cos ∅
cos ∅=1/sec ∅------> cos ∅=-5/13
sin ∅ is positive and cos ∅ is negative
so
∅ belong to the II quadrant
therefore
<span>the coordinates of point (x,y) on the terminal ray of angle theta are
</span>x=-5
y=12
the answer ispoint (-5,12)
see the attached figure
We'll use standard labeling of right triangle ABC, C=90 degrees, legs a, b, hypotenuse c.
11.
Right triangle, cliff peak A, boat B, angle opposite cliff is B=28.9 deg. adjacent leg a=65.7 m, cliff height is leg b.
tan B = b/a
b = a tan B = 65.7 tan 28.9° = 36.3 m
12.
Similar story, boat at B, opposite b=3.5 m, rope c=12 m
sin B = b/c
B = arcsin b/c = arcsin (3.5/12) = 17.0°
13.
c=124 m, A=58°
sin A = a/c
a = c sin A = 124 sin 58 = 105.2 m
14.
That's a hypotenuse c=4-1.2 = 2.8 m to a height b=1.8m so
cos A = b/c
A = arccos b/c = arccos (1.8/2.8) = 50.0°
15.
Not a right triangle, an isosceles triangle. Half of it is a right triangle with hypotenuse one arm, c=9.8 cm and angle opposite half the base of B=62/2=31°. We're after d=2b:
sin B = b/c
b = c sin B
d = 2b = 2 c sin B = 2(9.8) sin 31 = 10.1 cm
Almost equilateral
I Believe the mean would would 60 degrees Fahrenheit
Answer:
x = 26
y = 9
Step-by-step explanation:
(5x - 17)° + (3x - 11)° = 180° (angles in a straight line)
Solve for x
5x - 17 + 3x - 11 = 180
Collect like terms
5x + 3x - 17 - 11 = 180
8x - 28 = 180
Add 28 to both sides
8x = 180 + 28
8x = 208
Divide both sides by 8
x = 208/8
x = 26
Also:
(2y + 5)° + 90° + (3x - 11)° = 180° (angles on a straight line)
Plug in the value of x and solve for y
2y + 5 + 90 + 3(26) - 11 = 180
2y + 5 + 90 + 78 - 11 = 180
Collect like terms
2y + 162 = 180
Subtract 162 from both sides
2y = 180 - 162
2y = 18
y = 9 (dividing both sides by 2)
#2 10:20 of that helps can u help me on my question
