Find sin(a+B)

Mathematics

1 answer:

diamong [38]2 years ago

6 0

Answer:

Cos A=5/13

we have

Cos² A= $1-Sin ²A$

25/169=1-Sin²A

sin²A=1-25/169

sin²A=144/169

Sin A= $\sqrt{144/169}=12/13$

again

Tan B=4/3

P/b=4/3

p=4

b=3

h= $\sqrt{3²+4²}=5$

Now

Sin B=p/h=4/5

in IV quadrant sin angle is negative so

Sin B=-4/5

CosB=b/h=3/5

Now

Sin(A+B):sinAcosB+CosAsinB

now 

substitute value

Sin(A+B):12/13*3/5+5/13*(-4/5)=36/65-4/13

=16/65 is a required answer

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Y’all I really need help with this, please help !!

gogolik [260]

2. The chord is bisected by the segment from the center, so you have a right triangle with legs 4 and 6. The hypotenuse is the radius of the circle, x.
x = √(4^2 +6^2) = √(16+36) = √52
x = 2√13

4. By the rules for secants, the product of the lengths of the parts of the chord is the same in each case.
3*7 = 2*x
x = 21/2
x = 10.5

6. The measure of angle "a" is the average of the two intercepted arcs.
a = (165° + 55°)/2
a = 110°
Of course, b is its supplement.
b = 180° - 110°
b = 70°

8. Secant rules again. The products of the near and far lengths are the same in each case. When the segment is a tangent, the near and far lengths are the same, so it is the square of the tangent length.
x² = 4(4+12) = 64
x = 8

10. This works the same way as problem 2. Here, it looks like you have a triangle with hypotenuse 6 and side length 3. The other side length is
half-chord = √(6² -3²) = √(36 -9) = √27 = 3√3
AB = 2*(half-chord) = 2*3√3
AB = 6√3

12. The equation of a circle of radius r and center (h, k) is
(x-h)² + (y-k)² = r²
Of course, the "passes through" point must satisfy the equation, so you have
(x-3)² + (y-0)² = r² = (-2-3)² + (-4-0)² = 25+16
(x-3)² + y² = 41

14. Same as 12. Radius is 4.
(x+5)² + (y-2)² = 16

16. Radius 2, center (0, -1).
x² + (y+1)² = 4

18. The angle bisectors for the axes are the lines
y = x
y = -x
Here's a plot.