ai93
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Just attempted another exam question. Would you mind correcting me if I am wrong?
The question is
A triangle ABC has sides of length AB= 3.5 m, BC = 5.1 m and AC = 4.2m.
a) Calculate the size of the angle B and the size of the angle C, in degrees correct to 1 decimal place, in each case.
b) Calculate the area of triangle ABC.
My solution
$$a^{2}=b^{2}+c^{2}-2bcCOSA$$
$$\therefore COSA=\frac{b^{2}+c^{2}-a^{2}}{2bc}$$
$$COSA=\frac{4.2^{2}+3.5^{2}-5^{2}}{2\cdot4.2\cdot3.5}$$
A$$=COS^{-1}(0.132)$$
A= 82.4
Sine Rule
$$\frac{sin(82.4)}{5.1}=\frac{SinB}{4.2}$$
SinB=$$\frac{sin(82.4)\cdot4.2}{5.1}$$
B=Sin$$^{-1}(0.816)
B=54.7$$
Now, 180-sin$$^{-1}(0.816)=125.3$$
So angle B can equal 54.7 or 125.3
Angle C
180-(82.4+54.7)=42.9<180
or
180-(82.4+125.3)=-27.6 (which cannot be possible)
$$\therefore$$C = 42.9 degrees
and, 42.9+82.4+54.7=180
Question b)
Area=$$\frac{1}{2}bccosC$$
$$\frac{1}{2}\cdot5.1\cdot4.2 \cdot cos(42.9)$$=7.8m
Hope you understand that, I put what you told me into practice and hopefully solved this question correctly.
The question is
A triangle ABC has sides of length AB= 3.5 m, BC = 5.1 m and AC = 4.2m.
a) Calculate the size of the angle B and the size of the angle C, in degrees correct to 1 decimal place, in each case.
b) Calculate the area of triangle ABC.
My solution
$$a^{2}=b^{2}+c^{2}-2bcCOSA$$
$$\therefore COSA=\frac{b^{2}+c^{2}-a^{2}}{2bc}$$
$$COSA=\frac{4.2^{2}+3.5^{2}-5^{2}}{2\cdot4.2\cdot3.5}$$
A$$=COS^{-1}(0.132)$$
A= 82.4
Sine Rule
$$\frac{sin(82.4)}{5.1}=\frac{SinB}{4.2}$$
SinB=$$\frac{sin(82.4)\cdot4.2}{5.1}$$
B=Sin$$^{-1}(0.816)
B=54.7$$
Now, 180-sin$$^{-1}(0.816)=125.3$$
So angle B can equal 54.7 or 125.3
Angle C
180-(82.4+54.7)=42.9<180
or
180-(82.4+125.3)=-27.6 (which cannot be possible)
$$\therefore$$C = 42.9 degrees
and, 42.9+82.4+54.7=180
Question b)
Area=$$\frac{1}{2}bccosC$$
$$\frac{1}{2}\cdot5.1\cdot4.2 \cdot cos(42.9)$$=7.8m
Hope you understand that, I put what you told me into practice and hopefully solved this question correctly.