1. ## 3D Trig

From point C, an observer on a yacht measures the angle of elevation of point D at the top of a cliff to be 25 degrees. At this point the bearing of the cliff from the yacht is 300 degrees. After the yacht has sailed 100 m on a fixed course to pt A, the cliff is due north of the observer and the angle of elevation to the top of the cliff is 15 degrees. Point B is vertically below D at the base of the cliff.
a) Show that angle ABC = 60 degrees.
b) Show that BC = BDtan65 and that AB = BDtan75
c) Calculatethe height of the cliff.
I've done parts a + b. Could someone show me how to do c). The answer is 31 m but I can't seem to get it. I used Pythagoras' Theorem.

2. Originally Posted by xwrathbringerx
From point C, an observer on a yacht measures the angle of elevation of point D at the top of a cliff to be 25 degrees. At this point the bearing of the cliff from the yacht is 300 degrees. After the yacht has sailed 100 m on a fixed course to pt A, the cliff is due north of the observer and the angle of elevation to the top of the cliff is 15 degrees. Point B is vertically below D at the base of the cliff.
a) Show that angle ABC = 60 degrees.
b) Show that BC = BDtan65 and that AB = BDtan75
c) Calculatethe height of the cliff.
I've done parts a + b. Could someone show me how to do c). The answer is 31 m but I can't seem to get it. I used Pythagoras' Theorem.
Triangle ABC is not a right angled triangle so Pytrhogaras' theorem doesn't apply. Try the cosine rule (you should end up with a quadratic equation with BD as your variable.)

3. Originally Posted by xwrathbringerx
From point C, an observer on a yacht measures the angle of elevation of point D at the top of a cliff to be 25 degrees. At this point the bearing of the cliff from the yacht is 300 degrees. After the yacht has sailed 100 m on a fixed course to pt A, the cliff is due north of the observer and the angle of elevation to the top of the cliff is 15 degrees. Point B is vertically below D at the base of the cliff.
a) Show that angle ABC = 60 degrees.
b) Show that BC = BDtan65 and that AB = BDtan75
c) Calculatethe height of the cliff.
I've done parts a + b. Could someone show me how to do c). The answer is 31 m but I can't seem to get it. I used Pythagoras' Theorem.
hi

since triangle ABC has its sides and one of its angle known , we can use the cosine rule with the help the answers obtained in (a) and (b)

$100^2=(BD\tan 75)^2+(BD\tan 65)^2-2(BD\tan 75)(BD\tan 65)\cos 60$

solving that , BD=31 m and it doesn't agree with the answer provided .

hi

since triangle ABC has its sides and one of its angle known , we can use the cosine rule with the help the answers obtained in (a) and (b)

$100^2=(BD\tan 75)^2+(BD\tan 65)^2-2(BD\tan 75)(BD\tan 65)\cos 60$

solving that , BD=53.3 m and it doesn't agree with the answer provided .
Yes it does give the answer approx 31m as provided. Mathaddict - I think something went wrong with your calculations.

6. Originally Posted by xwrathbringerx
I'm not sure what you both are doing but I just did it again and got 31.
It comes down to:
10000 = 10.5237 x BD^2
Did you get 10.5237 when you add the coefficients?

7. Originally Posted by Debsta
I'm not sure what you both are doing but I just did it again and got 31.
It comes down to:
10000 = 10.5237 x BD^2
Did you get 10.5237 when you add the coefficients?
Debsta is right . Sorry for creating this confusion .

xwrathbringerx , try arranging your parantheses properly , then you should get 31 from the equation i posted earlier.