+0

# These Questions are also due in 2 days

0
203
5

Find AC in the diagram to the nearest integer. $$In \triangle PQR, we have \angle P = 90^\circ, PQ=3, and QR=\sqrt {58}. Find \tan R.$$ $$Find the value of \sin 40^\circ \cdot \cos 50^\circ + \cos 40^\circ \cdot \sin 50^\circ.$$(Says: "Find the value of...")

In the diagram below, we know $$\tan \theta = \frac{3}{4}.$$ Find the area of the triangle. May 30, 2019

#1
+1

Nvm, the last one I got it. It is 720

May 30, 2019
#2
+3

Find AC in the diagram to the nearest integer. sin( angle )  =  opposite / hypotenuse

sin( 70° )  =  31 / AC

Multiply both sides of the equation by  AC

AC sin( 70° )  =  31

Divide both sides of the equation by  sin( 70° )

AC  =  31 / sin( 70° )

Plug    31 / sin( 70° )    into a calculator.

AC  ≈  33

May 30, 2019
#3
+3

In $$\triangle PQR,$$ we have $$\angle P = 90^\circ,$$ $$PQ=3,$$ and $$QR=\sqrt {58}.$$ Find $$\tan R.$$ By the Pythagorean theorem,

32 + ( PR )2  =  √[ 58 ]2

9  +  ( PR )2  =  58

( PR )2  =  49

PR  =  7

tan( angle )  =  opposite / adjacent

tan( R )  =  3 / PR

tan( R )  =  3 / 7

May 30, 2019
#4
+3

Find the value of   $$\sin40°\cdot\cos50°+\cos40°\cdot\sin50°$$ .

Remember the angle addition formula for sine:    $$\sin(\alpha+\beta)\ =\ \sin\alpha\cos\beta+\cos\alpha\sin\beta$$

So...

$$\sin40°\cdot\cos50°+\cos40°\cdot\sin50°\ =\ \sin(40°+50°)\ =\ \sin(90°)\ =\ 1$$

_ _ _ _ _

May 30, 2019
#5
+3

Part of this last problem is just like the second problem.......

In the diagram below, we know  $$\tan\theta=\frac34$$ . Find the area of the triangle.

Let the base of the triangle be the side with length 60. Draw a height to that base and call it  h . By the Pythagorean Identity,

 1+ cot2θ  =  csc2θ 1 + ( 4/3 )2  =  1 / sin2θ 25 / 9  =  1 / sin2θ 9 / 25  =  sin2θ sin θ  =  3 / 5

 sin( angle )  =  opposite / hypotenuse sin θ  =  h / 40 3 / 5  =  h / 40 24  =  h

 area of triangle  =  (1/2)(base)(height) area of triangle  =  (1/2)(60)(24) area of triangle  =  720
May 30, 2019