Please help with maths homework?
how do you do this question
Solve triangle PQR. Give angles to the nearest whole second. Check the result.
p = 18
q = 19
r = 20
Answers:
You can use the law of cosines, which says that:
(c^2 - a^2 -b^2)/(2*a*b) = cosC
where a, b, and c are the lengths of the sides of the triangle, and A, B, C are the angles opposide the sides a,b,c, respectively. By picking different sides as "side c", you can find all the angles in your problem.
I get that:
angle R (the one opposite side r) is 65.37568 degrees = 65deg 22 min 32 sec
angle Q (the one opposite side q) is 59.72395 degrees = 59 deg 43 min 26 sec
and angle P (the one opposite side p) is 54.90037 degrees = 54 deg 54 min 1 sec
See source for an on-line calculator that will carry out the calculations for you.
i dnt have tym 2 solve it now sorry!
s=(p+q+r)/2=(18+19+20)/2=28.5
tanA/2=[(s-b)(s-c)/s(s-a)]^1/2.
tanB/2=[(s-c)(s-a)/s(s-b)]^1/2.
once A/2 and B/2 are known A and B can be found out and C=180-(A+B)
can u please stop asking homework questions.
are u dumb?
do ur own homework!!!!!!!!!!!.
squiggly~~~~~~~~~~~~~~~~~~~~~~.
A triangle has 180 degrees.
Angles and sides must be in proportion to each other.
q is 5% smaller then r
p is 10% smaller than r
r + .95r +.9r = 180
2.85r=180
r=63.15789474
q=60
p=56.84210526
They give you three sides to a triangle and expect you to find angles.
Since you don't know if it's a right triangle, start with the law of cosines:
p^2 = q^2 + r^2 - 2*q*r*cos P (P is the angle between side q and side r, the angle opposite side p)
Solve for P
Find angle R next by using the law of sines:
sin P / p = sine R / r
Since you know p, r and P, you can solve for R.
Now add P and R together and subtract from 180 degrees and you get angle Q.
There's the triangle.
Shouldn't you be doing your own homework? No wonder kids are starting employment without basic writing and arithmatic skills.
use cosine rule
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Solve triangle PQR. Give angles to the nearest whole second. Check the result.
p = 18
q = 19
r = 20
Answers:
You can use the law of cosines, which says that:
(c^2 - a^2 -b^2)/(2*a*b) = cosC
where a, b, and c are the lengths of the sides of the triangle, and A, B, C are the angles opposide the sides a,b,c, respectively. By picking different sides as "side c", you can find all the angles in your problem.
I get that:
angle R (the one opposite side r) is 65.37568 degrees = 65deg 22 min 32 sec
angle Q (the one opposite side q) is 59.72395 degrees = 59 deg 43 min 26 sec
and angle P (the one opposite side p) is 54.90037 degrees = 54 deg 54 min 1 sec
See source for an on-line calculator that will carry out the calculations for you.
i dnt have tym 2 solve it now sorry!
s=(p+q+r)/2=(18+19+20)/2=28.5
tanA/2=[(s-b)(s-c)/s(s-a)]^1/2.
tanB/2=[(s-c)(s-a)/s(s-b)]^1/2.
once A/2 and B/2 are known A and B can be found out and C=180-(A+B)
can u please stop asking homework questions.
are u dumb?
do ur own homework!!!!!!!!!!!.
squiggly~~~~~~~~~~~~~~~~~~~~~~.
A triangle has 180 degrees.
Angles and sides must be in proportion to each other.
q is 5% smaller then r
p is 10% smaller than r
r + .95r +.9r = 180
2.85r=180
r=63.15789474
q=60
p=56.84210526
They give you three sides to a triangle and expect you to find angles.
Since you don't know if it's a right triangle, start with the law of cosines:
p^2 = q^2 + r^2 - 2*q*r*cos P (P is the angle between side q and side r, the angle opposite side p)
Solve for P
Find angle R next by using the law of sines:
sin P / p = sine R / r
Since you know p, r and P, you can solve for R.
Now add P and R together and subtract from 180 degrees and you get angle Q.
There's the triangle.
Shouldn't you be doing your own homework? No wonder kids are starting employment without basic writing and arithmatic skills.
use cosine rule
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