Part (a)
Focus on triangle PSQ. We have
angle P = 52
side PQ = 6.8
side SQ = 5.4
Use of the law of sines to determine angle S
sin(S)/PQ = sin(P)/SQ
sin(S)/(6.8) = sin(52)/(5.4)
sin(S) = 6.8*sin(52)/(5.4)
sin(S) = 0.99230983787513
S = arcsin(0.99230983787513)
S = 82.889762826274
Which is approximate
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Use this to find angle Q. Again we're only focusing on triangle PSQ.
P+S+Q = 180
Q = 180-P-S
Q = 180-52-82.889762826274
Q = 45.110237173726
Which is also approximate.
A more specific name for this angle is angle PQS, which will be useful later in part (b).
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Now find the area of triangle PSQ
area of triangle = 0.5*(side1)*(side2)*sin(included angle)
area of triangle PSQ = 0.5*(PQ)*(SQ)*sin(angle Q)
area of triangle PSQ = 0.5*(6.8)*(5.4)*sin(45.110237173726)
area of triangle PSQ = 13.0074347717966
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Next we'll use the fact that RS:SP is 2:1.
This means RS is twice as long as SP. Consequently, this means the area of triangle RSQ is twice that of the area of triangle PSQ. It might help to rotate the diagram so that line PSR is horizontal and Q is above this horizontal line.
We found
area of triangle PSQ = 13.0074347717966
So,
area of triangle RSQ = 2*(area of triangle PSQ)
area of triangle RSQ = 2*13.0074347717966
area of triangle RSQ = 26.0148695435932
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We're onto the last step. Add up the smaller triangular areas we found
area of triangle PQR = (area of triangle PSQ)+(area of triangle RSQ)
area of triangle PQR = (13.0074347717966)+(26.0148695435932)
area of triangle PQR = 39.0223043153899
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<h3>Answer: 39.0223043153899</h3>
This value is approximate. Round however you need to.
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Part (b)
Focus on triangle PSQ. Let's find the length of PS.
We'll use the value of angle Q to determine this length.
We'll use the law of sines
sin(Q)/(PS) = sin(P)/(SQ)
sin(45.110237173726)/(PS) = sin(52)/(5.4)
5.4*sin(45.110237173726) = PS*sin(52)
PS = 5.4*sin(45.110237173726)/sin(52)
PS = 4.8549034284642
Because RS is twice as long as PS, we know that
RS = 2*PS = 2*4.8549034284642 = 9.7098068569284
So,
PR = RS+PS
PR = 9.7098068569284 + 4.8549034284642
PR = 14.5647102853927
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Next we use the law of cosines to find RQ
Focus on triangle PQR
c^2 = a^2 + b^2 - 2ab*cos(C)
(RQ)^2 = (PR)^2 + (PQ)^2 - 2(PR)*(PQ)*cos(P)
(RQ)^2 = (14.5647102853927)^2 + (6.8)^2 - 2(14.5647102853927)*(6.8)*cos(52)
(RQ)^2 = 136.420523798282
RQ = sqrt(136.420523798282)
RQ = 11.6799196828694
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We'll use the law of sines to find angle R of triangle PQR
sin(R)/PQ = sin(P)/RQ
sin(R)/6.8 = sin(52)/11.6799196828694
sin(R) = 6.8*sin(52)/11.6799196828694
sin(R) = 0.4587765387107
R = arcsin(0.4587765387107)
R = 27.3081879220073
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This leads to
P+Q+R = 180
Q = 180-P-R
Q = 180-52-27.3081879220073
Q = 100.691812077992
This is the measure of angle PQR
subtract off angle PQS found back in part (a)
angle SQR = (anglePQR) - (anglePQS)
angle SQR = (100.691812077992) - (45.110237173726)
angle SQR = 55.581574904266
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<h3>Answer: 55.581574904266</h3>
This value is approximate. Round however you need to.
is the slope of the line
is the y intercept