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
Just add:
3 3/4+4 5/8=3 6/8+4 5/8= 8 3/8
then subtract:
11 1/8 - 8 3/8= 2 3/4
so she had 2 3/4 more that the other two
it will be d i hope this helps have a good day!!!
4. x^2 = 49 so x = +/- 7
5. 3x^2 = 27
x^2 = 9
x = +- 3
6.x^2 = 225
x = +/- 15
