Math Is Fun Forum
  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

#1 2006-12-04 07:33:41

Neha
Member
Registered: 2006-10-11
Posts: 173

Triangle

preschooltriangleyt7.gif

solve the triangle below for its unkown parts:

ok the unkown parts are the top of the triangle which is B degrees
the left side point of the triangle  is 65 degrees
right side point is C degrees

and
one side of the triangle the left side is 11inches
the other side of the triangle the right side is A inches
teh bottom of the triangle is 15inches

now i did:
65dg + C + B = 180
65dg + 65dg + B = 180
( i wrote 65dg for C angle because both sides have to be equal in my view. )
anyways
130 + B = 180
180 - 130 = 50
B = 50
C = 65
and now we have to find A side.

do like this :
11^2 + A^2 = 15^2
no right?

Last edited by Neha (2006-12-04 07:38:28)


Live Love Life

Offline

#2 2006-12-04 16:49:09

pi man
Member
Registered: 2006-07-06
Posts: 251

Re: Triangle

You made a couple of mistakes Neha.   First, you can't just assume angle C is 65 degrees just because the they look like they are in the picture.   Second, the Pythagoreom Theorem (a^2 + b^2 = c^2) only applies to right triangles.   

I think you're going to need some trig functions to solve this.   Here's what you know:
(Capital letters designate an angle, lowercase letters designate a side.   Side c would be opposite angle C).

A= 65 dg
B = ?
C = ?
a = ?
b = 15
c = 11

I could be totally wrong here since its been a long time since I studied trig, but I believe the Law of Cosines could be used to find the value of a.   Once you know that, the Law of Sines can be used to figure out B and C. 

The Law of Cosines followed by the Law of Sines:

I stated the Law of Cosines as it's normally written which makes it confusing in this case because you only know A.   So for you, it would be:

Hopefully this helps.   And if I'm wrong, hopefully somebody will correct me.

Offline

#3 2006-12-05 06:04:14

Neha
Member
Registered: 2006-10-11
Posts: 173

Re: Triangle

ok this is what i did:

c^2 = a^2 + b^2 - 2ab cos P
A^ = 11^2 + 15^2 - 2(11)(15)cos 65dg
A^2 = 121 + 225 - 139.4640264
A^2 = 346 - 139.4640264
A^2 = 206.5359736
A = 14.37135949
A = 14.4


NEXT

14.4/sin65 = 15/sinB
sinB = 15sin65/14.4
SinB = 0.945952038

NEXT

14.4/sin65 = 11/sinC
sinC = 11sin65/14.4
sinC = 0.012106812


CORRECT


Live Love Life

Offline

#4 2006-12-05 07:14:26

pi man
Member
Registered: 2006-07-06
Posts: 251

Re: Triangle

Looks good to me.   Just one small mistake and then one more step.   Your equation for calculating SIN(C) looks right but the answer is incorrect.  It should be SIN(C) = .692320138.

You now know what the SIN(B) and SIN(C) are equal to so you can use the arcsin function to determine the angle.

B = arcsin(.945952038) = 71.07
C = arcsin(.692320138) = 43.81

Add those 2 together plus your 65 for Angle A and you get 179.88.   Not quite 180 but I think that can be explained to rounding errors.

Offline

Board footer

Powered by FluxBB