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Question: In a triangle ABC , sinA , sinB , sinC are in AP then The altitudes are in AP The altitudes are ...

In a triangle ABC , sinA , sinB , sinC are in AP then
The altitudes are in AP
The altitudes are in HP
The angles are in AP
The angles are in HP

Explanation

Solution

Hint - In such questions representing all the three sides a , b , c in terms of a constant like area of the triangle would be used to find the relation between all the sides . Using what is given in the question with the sine formula would give you the desired result .

Complete step-by-step answer:


Let us consider the triangle ABC with sides a, b, c and altitude h1,h2,h3{h_1},{h_2},{h_3} upon the base BC , AC , AB respectively .
Let the area of the triangle ABC be R
R = 12\dfrac{1}{2} (h1×a)\left( {{h_1} \times a} \right) = 12\dfrac{1}{2} (h2×b)\left( {{h_2} \times b} \right) = 12\dfrac{1}{2} (h3×c)\left( {{h_3} \times c} \right)
( area of the triangle = 12\dfrac{1}{2} base ×\times height )
\Rightarrow a=2Rh1;b=2Rh2;c=2Rh3a = \dfrac{{2R}}{{{h_1}}};b = \dfrac{{2R}}{{{h_2}}};c = \dfrac{{2R}}{{{h_3}}}

Now using the sine formula
asinA=bsinB=csinC\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}} = k (constant)
\Rightarrow sinA = ak\dfrac{a}{k} ; sinB = bk\dfrac{b}{k} ; sinC = ck\dfrac{c}{k}
$$$$
Now according to the question sinA , sinB , sinC are in AP
\Rightarrow ak,bk,ck\dfrac{a}{k},\dfrac{b}{k},\dfrac{c}{k} are in AP

Now , putting values of a , b , c from above ,
1h12Rk;1h22Rk;1h32Rk\dfrac{1}{{{h_1}}}\dfrac{{2R}}{k};\dfrac{1}{{{h_2}}}\dfrac{{2R}}{k};\dfrac{1}{{{h_3}}}\dfrac{{2R}}{k} are in AP
\Rightarrow 1h1;1h2;1h3\dfrac{1}{{{h_1}}};\dfrac{1}{{{h_2}}};\dfrac{1}{{{h_3}}} are in AP ( cancelling out the constants )
Therefore ,
h1,h2,h3{h_1},{h_2},{h_3} are in HP
Altitudes are in HP .

Note -
In these questions it is suitable to find an indirect method rather than to directly solve what's given . Remember that each fraction in the Sine Rule formula should contain a side and its opposite angle. Note that you should try and keep full accuracy until the end of your calculation to avoid errors .
\Rightarrow 1h1;1h2;1h3\dfrac{1}{{{h_1}}};\dfrac{1}{{{h_2}}};\dfrac{1}{{{h_3}}} are in AP ( cancelling out the constants )
Therefore ,
h1,h2,h3{h_1},{h_2},{h_3} are in HP
Altitudes are in HP .