Question

Trigonometry questions

Answer 1

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