1. A triangular lot has two of its side measuring 50m and 40m, respectively. Find the length of the third side of the lot, if the two given sides from an angle of 120 degrees.

2. On a hill, inclined at an angle of 14.2 degrees with the horizontal stands a vertical tower. At a point, 62.5 m down the hill from the foot of the tower, the angle of elevation of the top of the tower is 43.6 degrees. How tall is the tower?

Question

1. A triangular lot has two of its side measuring 50m and 40m, respectively. Find the length of the third side of the lot, if the two given sides from an angle of 120 degrees.

2. On a hill, inclined at an angle of 14.2 degrees with the horizontal stands a vertical tower. At a point, 62.5 m down the hill from the foot of the tower, the angle of elevation of the top of the tower is 43.6 degrees. How tall is the tower?

anonymous · Answer

3. Two sides of a parallelogram have lengths of 15.6 cm and 33.0 cm. If one of the angles is 42.6 degrees, find (a) length of the shorter diagonal and (b) the area of the parallelogram.

anonymous · Answer

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anonymous · Answer

Use the cosine law.

anonymous · Answer

is the formula c^2= a^2+b^2-2ab cos C

anonymous · Answer

Yes but i prefer it this way:

$$CosC= \frac{ a ^{2}+ b ^{2} -c ^{2} }{ 2ab }$$

anonymous · Answer

so im gonna find first the angle of C?

anonymous · Answer

In this case, your c is your unknown and you have your angle, your a and your b so you can find c

anonymous · Answer

i got it. angle C= 120 degrees?

anonymous · Answer

We know the angle is 120. so the formula put together is:

$$Cos(120) = \frac{ 50^{2}+40^{2}-c ^{2} }{ 2 x 50 x 40 }$$

anonymous · Answer

Solve for c, I gtg now sorry.

anonymous · Answer

thanks

anonymous · Answer

????

anonymous · Answer

$$Cos(120) = \frac{ 2500 + 1600 - c ^{2} }{ 4000 }$$

anonymous · Answer

Can you solve for c now? and by c I mean the unknown side of the triangle.

anonymous · Answer

.69?

anonymous · Answer

no.....its 0.707

anonymous · Answer

No, Cross multiply! so the new equation becomes:

$$4000Cos(120) = 4100 - c ^{2}$$

Cos(120) = -0.5

-2000 = 4100-c^2
-6100 = -c^2
6100 = c^2

$$c = \sqrt{6100}$$

c =78.1m

anonymous · Answer

Want to do the second question now?

anonymous · Answer

yes :))

anonymous · Answer

On a hill, inclined at an angle of 14.2 degrees with the horizontal stands a vertical tower. At a point, 62.5 m down the hill from the foot of the tower, the angle of elevation of the top of the tower is 43.6 degrees. How tall is the tower?

anonymous · Answer

Just wait, I'll draw it first.

anonymous · Answer

ok :)

anonymous · Answer

attachment

anonymous · Answer

What is in red is for the orange triangle and what is in green is for the black triangle.

anonymous · Answer

Have you learnt SOH CAH TOA?

anonymous · Answer

yep

anonymous · Answer

Do SOH for the orange triangle to find the height of the tower.

anonymous · Answer

am i not gonna use the law of sine or law of cosine?

anonymous · Answer

Does it say to use those?

anonymous · Answer

We can if you want but SOH CAH TOA IS easier. Although if you're in that unit, we might as well use the sine law.

anonymous · Answer

not sure because, the problem is under the law of sine and cosine

anonymous · Answer

Ok, we'll use the sine law then :)

anonymous · Answer

ok wait.....ill try

anonymous · Answer

We know that a triangle's total angles = 180

The last angle for the orange triangle is 180 -90 - 43.6 = 46.4

anonymous · Answer

$$\frac{ a }{ \sin A } = \frac{ b }{ \sin B }$$

anonymous · Answer

OH!

anonymous · Answer

Let me se... She went DOWN the hill, dang!! I misread.

anonymous · Answer

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anonymous · Answer

That's the second one and... the follwoing will be the first.

anonymous · Answer

so it is not a right triangle?

anonymous · Answer

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