# Math Help - Finding the Height of a cliff

1. ## Finding the Height of a cliff

Hello again,

I have this question that I could not find the answer to:
At the point P, a ship observes that the angle of elevation of the cliff at point T is 32 degrees. It sails for 180m towards the cliff to reach the point Q and observes that the angle of elevation of the point T becomes 48 degrees. Calculate the height of the cliff. (TR)
How can one find the height of the cliff using trigonometric function if we only have the adjacent information of the angle 32 degrees...?

Crude diagram:

Thank you

2. Originally Posted by Lites
Hello again,

I have this question that I could not find the answer to:

How can one find the height of the cliff using trigonometric function if we only have the adjacent information of the angle 32 degrees...?

Crude diagram:

Thank you
Hi Lites,

you can use the two right angled triangles to find the angles at the top apex.
If you ignore the triangle with the 48 degree angle, then taking the larger one,
the top angle is 90-32=58 degrees.

Now taking the smaller one, the angle at the top is 90-48=42 degrees.

Therefore the top angle in the scalene triangle at the left of the smaller
right-angled triangle is 58-42=16 degrees.

Now you have an angle and it's opposite side in the scalene triangle,
so you can use the Sine Rule to find side lengths.

This will give you the hypotenuse length of the smaller right-angled triangle
by re-arranging

$\frac{hyp}{sin32}=\frac{180}{sin16}$

Once you have that length you can use $sin48$ to find the vertical height.

3. Originally Posted by Archie Meade
Hi Lites,

you can use the two right angled triangles to find the angles at the top apex.
If you ignore the triangle with the 48 degree angle, then taking the larger one,
the top angle is 90-32=58 degrees.

Now taking the smaller one, the angle at the top is 90-48=42 degrees.

Therefore the top angle in the scalene triangle at the left of the smaller
right-angled triangle is 58-42=16 degrees.

Now you have an angle and it's opposite side in the scalene triangle,
so you can use the Sine Rule to find side lengths.

This will give you the hypotenuse length of the smaller right-angled triangle
by re-arranging

$\frac{hyp}{sin32}=\frac{180}{sin16}$

Once you have that length you can use $sin48$ to find the vertical height.
Ow ok I get it so in the end I got 257.2cm as the vertical height.

But this:
$\frac{hyp}{sin32}=\frac{180}{sin16}$
What are you doing exactly here? Are you using the comparing similarities formula? Why sin 32 is used to find the hyp instead of using sin 16? (They're in one triangle right?) I'm sorry but I think I have not been taught by my teacher about this operation especially this involves non-right-angled triangle...

4. You could even more quickly find the angles in the scalene triangle PTQ
with 180-48=132 and 180-(132+32)=16.

You only have a side belonging to the scalene triangle, |PQ|=180m.
The point Q hasn't been labelled
You have all angles of the rightmost right-angled triangle QRT,
and the large external one PRT, and also the scalene one, PTQ.

To find the height, you need a side length of the rightmost right-angled triangle QRT,
or the larger right-angled triangle PRT, in order to use sin, cos or tan.

We don't have a side of either of these,
hence we must work with the scalene triangle.

this means using the sine or cosine rule, which solve non-right-angled triangles.

The sine rule is the easier one to use,
we use it when we have a side length and the angle opposite this side.
We have |PQ|=180m and angle PTQ=16 degrees.

$\frac{sinA}{a}=\frac{sinB}{b}$

$\frac{a}{sinA}=\frac{b}{sinB}$

where A and B are 2 angles in a non-right-angled triangle
(it works for right-angled triangles too of course! but we usually use
the standard definitions for sin, cos and tan with those)

and "a" is the side opposite A,
"b" is the length of the side opposite angle B.

5. Originally Posted by Archie Meade
You could even more quickly find the angles in the scalene triangle PTQ
with 180-48-132 and 180-(132+32)=16.

You only have a side belonging to the scalene triangle, |PQ|=180m.
The point Q hasn't been labelled
You have all angles of the rightmost right-angled triangle QRT,
and the large external one PRT, and also the scalene one, PTQ.

To find the height, you need a side length of the rightmost right-angled triangle QRT,
or the larger right-angled triangle PRT, in order to use sin, cos or tan.

We don't have a side of either of these,
hence we must work with the scalene triangle.

this means using the sine or cosine rule, which solve non-right-angled triangles.

The sine rule is the easier one to use,
we use it when we have a side length and the angle opposite this side.
We have |PQ|=180m and angle PTQ=16 degrees.

$\frac{sinA}{a}=\frac{sinB}{b}$

$\frac{a}{sinA}=\frac{b}{sinB}$

where A and B are 2 angles in a non-right-angled triangle
(it works for right-angled triangles too of course! but we usually use
the standard definitions for sin, cos and tan with those)

and "a" is the side opposite A,
"b" is the length of the side opposite angle B.
Ah now I understand the concept, thank you for explaining it...tomorrow I'll have my semester test...

6. Ok Lites,

good luck with that.

Remember

$sinA=\frac{opp}{hyp}$

$cosA=\frac{adj}{hyp}$

$tanA=\frac{opp}{adj}$

are defined for a right-angled triangle.

The sine rule and cosine rule work for all triangles,
but we don't typically use them for solving right-angled triangles
as we have the easier ways shown above,
along with Pythagoras' theorem

$opp^2+adj^2=hyp^2$

hence we tend to use the sine rule and cosine rule for non-right-angled triangles.

To decide which one of those to use....

If we can find a side and it's opposite angle, we can easily apply the sine rule
(because the sine rule is a little simpler), if we have a third dimension (side or angle).

If not, we typically use the cosine rule.

7. Originally Posted by Archie Meade
Ok Lites,

good luck with that.

Remember

$sinA=\frac{opp}{hyp}$

$cosA=\frac{adj}{hyp}$

$tanA=\frac{opp}{adj}$

are defined for a right-angled triangle.

The sine rule and cosine rule work for all triangles,
but we don't typically use them for solving right-angled triangles
as we have the easier ways shown above,
along with Pythagoras' theorem

$opp^2+adj^2=hyp^2$

hence we tend to use the sine rule and cosine rule for non-right-angled triangles.

To decide which one of those to use....

If we can find a side and it's opposite angle, we can easily apply the sine rule
(because the sine rule is a little simpler), if we have a third dimension (side or angle).

If not, we typically use the cosine rule.
Ow ok, I just want to sum it up and see if I understand what do you explain..

If we have right-angled triangle we usually use the trigonometric function (sin, cos, tan) in this form:

$sinA=\frac{opp}{hyp}$

$cosA=\frac{adj}{hyp}$

$tanA=\frac{opp}{adj}$

If we do not have right-angled triangle but we have a side and its opposite angle then we use the sine rule which is this:

Is that right?

But if we don't we have to use the cosine rule...now which form of cosine rule in this case?

Sorry if I asked too many questions..

8. The cosine rule is best left until the sine rule is mastered.

You can use it when you have the 3 dimensions of a corner

which are two sides and the angle between them,
and you use these 3 to find the side opposite the angle...

Or if you have all 3 sides you can find any angle

The cosine rule formula is

$a^2=b^2+c^2-2(bc)cosA$

$b^2=a^2+c^2-2(ac)cosB$

$c^2=a^2+b^2-2(ab)cosC$

Hence you need the dimensions of the opposite corner to find a side length

Rearranging the above we get

$2abCosC=a^2+b^2-c^2$

$CosC=\frac{a^2+b^2-c^2}{2ab}$

$C=arccos\left(\frac{a^2+b^2-c^2}{2ab}\right)$

We need to put all 3 sides into that formula to find C.

Similar story for A and B.

hence, to use the cosine rule you must be able to find at least 2 side lengths to begin with.

we cannot start your example using the cosine rule, therefore.
hence we had to use the sine rule anyway.

9. ## re: Finding the Height of a cliff

hello,
Antonia stands at the edge of a vertical cliff and throws a stone vertically upwards. The stone leaves her hand with a speed of 8.0 m/s. Acceleration due to gravity is 10 m/s2 and all distance measurements are taken from the point where the stone leaves Antonia's hand.
thanks!!
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