Question: If tan(A+B)=3 and tan(A−B)=31; 0°<A+B≤90°; A>B, find A and B.
Here, we are given that
tan(A+B)=3
tan(A−B)=31
0°<A+B≤90°
A>B
The most important part of this numerical is that we have to remember the values of tan for different angles, as shown in the table below
tan0°
tan30°
tan45°
tan60°
tan90°
0
31
1
3
Not defined
We know from the table above that, tan60°=3. So using this value in the 1st equation from given data:
tan(A+B)∴tan(A+B)∴A+B=3=tan60=60−−−(5)
Explanation for the above block:
Here we are comparing LHS with RHS. We can say that the equation tan (A+B) equates to √3 and tan 60 is √3 (you can find that in table).
So, this is only possible if A+B = 60 which satisfies this equation.
In the same manner, we know that, tan30°=31. So using this value in the 2nd equation from given data:
tan(A−B)=31∴tan(A−B)∴A−B=tan30=30−−−(6)
Adding equation (5) and equation (6), we get:
A+B=60A−B=30—————2A=90∴A=45°
Now, we can calculate the value of B by putting the value of A in either equation (5) or equation (6). Let's do it using equation (5):
A+B∴45+B∴B=15°=60=60
We can further confirm that our answer is correct by looking at equations (3) and equation (4) from the given data that:
0<AAnd so is 0<+B≤9060≤90
And,
A60>B>15
Hence, m∠A=60° and m∠B=15°
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Suggested Notes:
15 cot A = 8 | Find the value of sin A and sec A | Trigonometry Numerical
PQ = 12cm and PR = 13cm | Find tan P - cot R | Trigonometry Numerical
cot θ = 7/8 | Find all other trigonometric ratios | Trigonometry Numerical
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