triangle PQR lies on a horizontal plane and S is a point vertically above P?

Question

triangle PQR lies on a horizontal plane and S is a point vertically above P,PQ = 5cm , PR=10cm,RS=14cm and angleRPQ = 60degree.Find a)angle between plane QRS and PQR.......b)and between line PS and QR.

answer for a) 62.97 degree...b)90

Answer 1

Ok, It will be better to make a graphic, but, let's see, I am gonna give you everything step by step :

PS is perpendicular to the plane of the triangle PQR, and PR = 10 cm, PQ = 14, so, if the angle RPQ, is 60º, then QR will be :

QR^2 = 10^2 + 14^2 - 2*10*14*cos60

QR = sqrt(156) cm

Let's watch the triangle RPS, if PR = 10, and RS = 14, then

PS^2 = 14^2 - 10^2 = sqrt(96).

The angle between plane QRS and PQR, is form like this :

from S, draw a perpendicular line to QR, and from P, draw a perpendicular line to QR. So the both lines will touch on the same point, let's call the point A the angle form by those two lines, is the angle you are looking for.

You can find the value of PA, PA is the height of the triangle PQR.

the area of PQR = 14*10*sin(60)/2

but also, the area of PQR = base*height / 2 = sqrt(75)*h / 2

70sin(60) = sqrt(156)*h/2

h = 5 cm

then, the angle will be : tan(angle) = sqrt(96) / 5

angle = 62.9º

I got that answer

and the angle between both planes, is 90º, that's obvius

Answer 2

a) Find the dihedral angle between the horizontal plane containing triangle PQR and the inclined plane containing triangle QRS.

Given
PQ = 5 cm
PR = 10 cm
RS = 14 cm

S is directly above P.
Angle RPQ = 60°

First let's look at triangel PQR.

QR² = PQ² + PR² - 2(PQ)(PR) cos( QR² = 5² + 10² - 2*5*10*cos(60°)
QR² = 25 + 100 - 100*(1/2) = 125 - 50 = 75

QR = √75 = 5√3 cm

PQ = 5 cm
PR = 10 cm
QR = 5√3 cm

So triangle PQR is a right triangle with the 90° at Q. And since S is vertically above P, RPS is a right angle.

By the Pythagorean Theorem

PS² = RS² - PR² = 14² - 10² = 196 - 100 = 96
PS = √96

And since S is vertically above P, QPS is a right angle.

tan(
arctan(√96/5) ≈ 62.964308°

This angle is the same as the dihedral angle between the two planes.

b) Find the angle between line PS and QR.

These lines are skew but obviously from a) are perpendicular. The angle between them is 90°.

Answer 3

I'll give you a hint: use Pythagoras Theorem to find all sides for the vertical side of the triangle (treat it like a separate triangle): PS²+RS²=QS²
Fill in the information you have and then use your angle of 60° in formulas from trigonometry, like the law of sines, cosines, and so on, to fill in all the other angles and sides. You also know that since point PS is vertical, the angle below it must be 90° and the total of all angles in a triangle must = 180°.