i need to know the answers and how u got the answers for each of these pages i need to learn this missed lesson

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- Sep 2nd 2007, 11:54 AMsasha9193pyramid
i need to know the answers and how u got the answers for each of these pages i need to learn this missed lesson

- Sep 2nd 2007, 12:00 PMsasha9193
ok well question is....

a golfer plays a shot from t

the ball travels 120m to p

the hole (h) is 210m from the tee

angle HTP is 22degrees

i need to know distance from p to h

the angle pht

and the area of the triangle

need formulas as to how u do it and answers

so answer as if its said write your working out - Sep 2nd 2007, 12:05 PMsasha9193
heres a link to the work i dont understand need guidance on each question soz missed this total subject

MyMaths.co.uk - ow13Trigonometry - Sep 2nd 2007, 12:24 PMJhevon
**To the golf problem:**

for the first question, we can use the cosine rule (the Law of Cosines)

Just so you know:

$\displaystyle t$ is the side opposite angle $\displaystyle T$

$\displaystyle p$ is the side opposite angle $\displaystyle P$

$\displaystyle h$ is the side opposite angle $\displaystyle H$

we are given $\displaystyle p$ and $\displaystyle h$ and $\displaystyle T$ and we want to find $\displaystyle T$

By the cosine rule:

$\displaystyle t^2 = h^2 + p^2 - 2hp \cos T$

use this formula to find t.

for the second question, we can use the sine rule (the Law of Sines) now that we've found t

By the sine rule:

$\displaystyle \frac {t}{\sin T} = \frac {h}{\sin H}$

use this formula to solve for $\displaystyle H$

for the third question:

we can use the formula, $\displaystyle \mbox { Area } = \frac {1}{2}ph \sin T$ - Sep 2nd 2007, 12:30 PMsasha9193
im not getting the first bit can you rearrange the formula to work it out for me i wanna know how to do that

- Sep 2nd 2007, 12:31 PMsasha9193
cus i do it but get t still on 1 side and tsquared on the other, can you plz work it out with full calculations that will help me out alot

- Sep 2nd 2007, 12:51 PMJhevon
**To the Pyramid question:**

see the diagrams below. i drew in some lines and added some figures here. the red lines correspond to the first question, the pink lines correspond to the last question

first question: the told us E is above the midpoint, so if we drew in lines that cut the midpoint of AB at 90 degrees and the midpoint of BC at 90 degrees, they would meet at the midpoint of the base below E. I will call the midpoint o. Now see, we can construct a right-triangle, with hypotenuse of length = ? and legs of length 4 and 5.

we can find ? by using Pythagoras' Theorem.

**Pythagoras' Theorem:**If the hypotenuse of a right-triangle is a and the legs are b and c, then we have,

$\displaystyle a^2 = b^2 + c^2$

once we find ?, we can find another right-triangle AOE.

we know that the legs have length 13 and ? (which we found above), now we use Pythagoras' theorem again, to find AE

for the second question:

we can construct the right-triangle using the pink lines on the diagram. the legs are 13 and 4. we don't know the hypotenuse

we wish to find the angle x. we can use the tangent ratio to do this

remember, $\displaystyle \tan x = \frac {\mbox { Opposite }}{ \mbox { Adjacent }} = \frac {13}{4}$ in this case

so then $\displaystyle x = \tan^{-1} \left( \frac {13}{4}\right)$ - Sep 2nd 2007, 01:08 PMsasha9193
Ty But I Need Help On The Very First Question I Need The Answer And Full Working Out

- Sep 2nd 2007, 01:29 PMsasha9193
Ok I Gtg Bed And I Need This Done Soon Plz Go To That Site And Give Me Full Answers With Working Out So I Can Learn From That Tomorrow But I Really Need Answers Tonight Plz

FIRST ANSWERS TO THE QUESTION I FIRST POSTED THEN HOW I DO PAGES 2-4 ON THAT LINK WITH FULL ANSWERS PLZ I REALLY NEED IT NOW OR IM IN BIG TROUBLE - Sep 2nd 2007, 01:35 PMJhevon
- Sep 2nd 2007, 01:36 PMsasha9193
im really sorry but i was sick for ages

- Sep 2nd 2007, 01:37 PMsasha9193
i jus need first 3 answers for questions at very beginning of post then if u can go on the site and give me full answers for the pages 2-4 ty alot if u can u a life saver