Q1) Find area of the triangle ABC where AB=c, AC=b and BC=a. When a=b=c=12m
4 answers
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Area of tiangle = 1/2 base*altitude
= 1/2 *12*12
= 1/2 *12*12
Area of tiangle = 1/2 base*altitude
= 1/2 *12*12
= 72sq. unit
= 1/2 *12*12
= 72sq. unit
1. Label all the sides as a, b and c. Where A is 15m, B is 10m & C is 12m (as these are corresponding to their sides.
2. Write down the formula for the cosine rule (because were looking for the angle, CosA will be the subject rather than a^2):
CosA=b^2+c^2-a^2/2bc
2.Substitute your values:
CosA=10^2+12^2-15^2/2*10*12
4. Now, because you will be looking for the angle, you need to enter SHIFT + Cos on your calculator in order for it find the angle.
5. Your final answer should be 85.5 degrees (Rounded to 1 d.p.)
2. Write down the formula for the cosine rule (because were looking for the angle, CosA will be the subject rather than a^2):
CosA=b^2+c^2-a^2/2bc
2.Substitute your values:
CosA=10^2+12^2-15^2/2*10*12
4. Now, because you will be looking for the angle, you need to enter SHIFT + Cos on your calculator in order for it find the angle.
5. Your final answer should be 85.5 degrees (Rounded to 1 d.p.)