Determine the length of the equal sides of the triangle...

[confused-]

If you want to draw something, go to http://www.twiddla.com/137515 and draw stuff.

The length of a base in an isosceles triangle is 10cm. Determine the length of the equal sides of the triangle if each make an angle of 50deg with the base.

This is in Chapter 7 called Right-Angled Trigonometry, except it doesn't have a right angle.

I tried this and it didn't work:

tan 50deg = O/A = x/10 = 11.92

Edit: I tried cos 50 = a/h = 10/x
x*(cos 50 ) = 10
x= 10/cos 50
=15.56

The answer at the back is 7.78m if it helps.

Edit: I have some more questions...

Evaluate each of the following, correct to the nearest minute

11 a) sin[sup:3pbbk75r]-1[/sup:3pbbk75r] 0.1572

I put it in my calculator and it says this:
9' 2' 39.88'

Answer is this:
9' 03'



...and last ones (2 questions in picture):

15 & 16)

For 15 I have figured out s but not t.

For t I have:

tan 68deg = O/A = s/t = 41.69/t
tan 68t = 41.69
68t = tan[sup:3pbbk75r]-1[/sup:3pbbk75r] 41.69
t= 1.30

But I'm way off, the answer is 16.84

For 16 I have this for alpha :

sin alpha = o/h = 79/82

but the answer is 74' 27' and when I put 79/82 into the calculator i get 0' 57'

I'm not sure about beta because it doesn't say that there is another right angle next to the other right angle.

 

[mmm4444bot]

… [this exercise] doesn't have a right angle. Yes it does, once you know where to look.

Draw a picture of some arbitrary isosceles triangle; orient your drawing so that the 10-centimeter base is on the bottom.

Next, cut the triangle in half from top to bottom. In other words, draw a line from the middle of the 10-centimeter side to the opposite vertex.

Now you have two identical right-triangles.

Use either one of them to answer the exercise.

 

[confused-]

Re:

… [this exercise] doesn't have a right angle. Yes it does, once you know where to look.

Draw a picture of some arbitrary isosceles triangle; orient your drawing so that the 10-centimeter base is on the bottom.

Next, cut the triangle in half from top to bottom. In other words, draw a line from the middle of the 10-centimeter side to the opposite vertex.

Now you have two identical right-triangles.

Use either one of them to answer the exercise.

Oh! :idea:

 

[Denis]

For 15 I have figured out s but not t.

Next time, PLEASE start a NEW thread for a new problem.

t / SIN(22) = s / SIN(68)

 

[mmm4444bot]

Evaluate each of the following, [rounded] to the nearest minute

11 a) sin[sup:1rvicgsq]-1[/sup:1rvicgsq] 0.1572

I put it in my calculator and it says this:

9' 2' 39.88'

Answer is this, and it's rounded to the nearest minute:

9' 03'

Do you understand how to round 39.88 seconds to the nearest minute?

tan 68deg = O/A = s/t = 41.69/t

tan(68 deg) is a constant!

Continue with that.

2.4751 = 41.69/t

For 16 I have this for alpha :

sin alpha = o/h = 79/82

but the answer is 74' 27' and when I put 79/82 into the calculator i get 0' 57' Put 79/82 into the inverse sine function first; you're looking for an angle, right?

Trig ratios go into inverse functions, and angles come out.

I'm not sure about beta because it doesn't say that there is another right angle next to the other right angle.

Are you kidding?!

Of course they are both right angles. :?

cos(?) = ADJ/HYP