PLEASE HELP! THIS IS PAST DUE AND I HAVE NO IDEA HOW TO DO IT! 10 GRADE MATH

Question

3. Given a right triangle DEF with angle D as a right angle, find side DE if DF = 5 units and EF = 13 units.

4. Given a right triangle PQR with angle R as 90 degrees and angle Q as 54 degrees, find side PR if side RQ is 4.6 units.

5. Given a right triangle ABC with angle C as 90 degrees, find angle A if side AC = 3.0 units and BC = 6.8 units and AC = 3.0 units, find angle A.

Answer 1

Pythagorean equation:

a^2 + b^2 = c^2

D is the right angle (90°),

DE can be considered side a and DF can be consider side b and EF is side c (the hypotenuse)

Plug into equation and solve for DE

Have you studied trigonometric equations:

cos of angle = adjacent side / hypotenuse

sin of angle = opposite side / hypotenuse

tan of angle = opposite side / adjacent side

Draw right triangle,

label right angle R, the vertical side from R could be PR and the horizontal side from R could be QR

Standing at the Q corner .... we see side PR opposite to us and QR adjacent (next to) to us

So, which trigonometric equation has those components?

Plug angle into calculator and perform operation, then do algebra to solve for PR

Have you studied the Law of Sines and/or Law of Cosines:

Law of Sines indicates that: length of A / sin A = length of B / sin B = length of C / sin C

Draw right triangle,

label right angle C, the vertical side from C could be AC and the horizontal side from C could be BC

Use Pythagorean equation to solve for hypotenuse AB

Plug in length of A / sin A = hypotenuse / sin 90°

Do some rearranging to solve for sin A and take the inverse of sin on calculator to get angle A