Finding the Base Angles of a Triangle

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Problem Corner

C

^{onsider a}^△ABCin which the following are

specified: ∡A(i.e., the apex angleBAC), the length aof the baseBC, and the lengthhof the altitude fromAtoBC.

Is it possible to find expressions for the two base angles,∡B and∡C, in terms ofA,a,h? We do so using trigonometry.

Figure 1.

LetADbe the perpendicular from vertexAtoBC, and let BD=x,DC=a−x. (For convenience, we assume that∡B and∡Care acute, which means thatDlies on the side and not on the extension of the side. We also assume that the triangle is not right-angled.)

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Keywords: trigonometry, quadratics, problem solving

Finding the Base

Angles of a Triangle

MOSES MAKOBE

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68 Azim Premji University At Right Angles, March 2022

From right-angled trianglesABDandACD, we have:

tanB= ^h

x, tanC= ^h

a−x. (1)

Hence:

x= ^h

tanB, a−x= ^h

tanC =−_tan₍_A^h₊_B₎, (2)

where the last step comes from the fact thatC=180^◦−(A+B). Hence:

a

h = ^tan⁽^A⁺^B⁾⁻^tan^B

tanB·tan(A+B)

=( _tan_A₊_tan_B

1−tanA·tanB−tanB)

÷tanB·( _tan_A₊_tan_B

1−tanA·tanB

)

= ^tan^A⁺^tan^A^·^tan²^B

tanB·(tanA+tanB).

From the last relation we obtain, by cross-multiplication:

(a−htanA)tan²B+ (atanA)tanB−htanA=0. (3) Here, (3) can be regarded as a quadratic equation in tanB; the coefficients are known quantities, as they have been expressed in terms ofa,h,A. Solving the equation, we get:

tanB= ⁻^a^tan^A^±

√a²tan²A+4(a−htanA)·htanA

2(a−htanA) . (4)

We have not attempted to simplify the expression in (4). An equivalent way of expressing the same result, in terms of sines and cosines, is the following:

tanB= ^−a^sin^A^±

√a²sin²A+4(acosA−hsinA)·hsinA

2(acosA−hsinA) . (5)

Note the plus-minus sign. The two values given by the formula correspond to the values of tanBand tanC respectively. (There is an obvious symmetry in the problem betweenBandC.)

If the triangle is right-angled, then we may encounter fractions with zero denominator, so we need to be careful. We look at this possibility below.

The case when A=90^◦.In this case,∡B+∡C=90^◦, so tanB·tanC= 1. Therefore (2) assumes the form

x= ^h

tanB, a−x= ^h

tanC =htanB, ∴ x(a−x) =h². (6) The quadratic equationx(a−x) =h²may be solved forx, and from this we get tanB:

x(a−x) =h², ∴ x= ^a^±

√a²−4h²

2 ,

∴ tanB= ^2h

a±√

a²−4h². (7)

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Rationalising, we get:

tanB= ^a^∓

√a²−4h²

2h . (8)

As earlier, the two values given by the formula correspond to the values of tanBand tanCrespectively.

(Note that the product of the two values is equal to 1, as it should be.)

The case when a denominator of 0 occurs in(4)and(5).This will happen whena=htanA (equivalently,acosA=hsinA). This means, clearly, that either∡B=90^◦or∡C=90^◦. References

1. Bradley M, et al, 2012,Platinum mathematics grade11learner’s book, Cape Town, Maskew Miller Longman 2. Aird J, et al, 2013,Clever Keeping Maths Simple grade12learner’s book, Northlands, MacMillan South Africa

LETUKU MOSES MAKOBE is the HOD for Faculty of Sciences at Makwe Senior Secondary School, Limpopo province, RSA. He teaches Mathematics and Physical sciences for grades 10-12. He is the founder of the project

`Makobe Mathematics Club’ which provides free lessons in mathematics and the sciences to underprivileged learners. He holds a post graduate diploma in public management from Dr. C N Phatudi College of Education, UNISA (CIMSTE), Regenesys Business School. He is an active contributor of articles to AMESA. He may be contacted at [email protected].