Question

In a triangle ABC, AD is the altitude from A, b>c. \[C = {23^ \circ }\], \[AD = \dfrac{{abc}}{{{b^2} - {c^2}}}\]. Then B is equal to:
               

A.\[{70^ \circ }\]
B. \[{113^ \circ }\]
C. \[{123^ \circ }\]
D. \[{103^ \circ }\]

Answer 1

Hint: Here we use the method of perpendicular divided by hypotenuse to write the value of sine of an angle in a right triangle. Equate the values of AD obtained from the sine formula and from the statement in the question. Use the law of sine to write the relation between angles A and C. Substitute the values of sides using the law of sine and calculate the angle B.
* Law of sine states that in a triangle ABC the ratio of side of a triangle to the sine of opposite angle is same for all the sides of the triangle, i.e. if side a has opposite angle A, side b has opposite angle B and side c has opposite angle C, then we can write \[\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c} = k\]where k is some constant term.
* In a right triangle, if one of the angle is \[\theta \]then;
\[\sin \theta = \]Perpendicular divided by hypotenuse
\[\cos \theta = \]Base divided by hypotenuse

Complete step-by-step answer:
Since, AD is the altitude from A, then \[\angle ADC = {90^ \circ }\].
               

So, in the right triangle ADC, we know \[C = {23^ \circ }\].
\[\sin {23^ \circ } = \dfrac{{AD}}{{AC}}\]
Substitute the value of AC as b
\[ \Rightarrow \sin {23^ \circ } = \dfrac{{AD}}{b}\]
Cross multiply the terms
\[ \Rightarrow AD = b\sin {23^ \circ }\]
Also, we are given in the statement of the question that \[AD = \dfrac{{abc}}{{{b^2} - {c^2}}}\].
Equate both the values of AD.
\[ \Rightarrow \dfrac{{abc}}{{{b^2} - {c^2}}} = b\sin {23^ \circ }\]
Cancel the same terms from both sides of the equation.
\[ \Rightarrow \dfrac{{ac}}{{{b^2} - {c^2}}} = \sin {23^ \circ }\] … (1)
Now in triangle ABC we use the law of sine.
\[\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c} = k\]
Take the pair of sides a and c
\[ \Rightarrow \dfrac{{\sin A}}{a} = \dfrac{{\sin C}}{c}\]
Substitute the value of angle \[C = {23^ \circ }\]
\[ \Rightarrow \dfrac{{\sin A}}{a} = \dfrac{{\sin {{23}^ \circ }}}{c}\]
Substitute the value of \[\sin {23^ \circ }\]from equation (1)
\[ \Rightarrow \dfrac{{\sin A}}{a} = \dfrac{{\dfrac{{ac}}{{{b^2} - {c^2}}}}}{c}\]
Write the fraction in RHS in proper form
\[ \Rightarrow \dfrac{{\sin A}}{a} = \dfrac{{ac}}{{({b^2} - {c^2})c}}\]
Cancel the same terms from numerator and denominator.
\[ \Rightarrow \dfrac{{\sin A}}{a} = \dfrac{a}{{({b^2} - {c^2})}}\]
Cross multiply the term in the denominator of LHS to the numerator of RHS.
\[ \Rightarrow \sin A = \dfrac{{{a^2}}}{{{b^2} - {c^2}}}\] … (2)
Now we know from law of sine \[\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c} = k\]
Squaring all the terms we get \[\dfrac{{{{\sin }^2}A}}{{{a^2}}} = \dfrac{{{{\sin }^2}B}}{{{b^2}}} = \dfrac{{{{\sin }^2}C}}{{{c^2}}} = {k^2}\]
Therefore, values after cross multiplying the terms with k will be
\[{a^2} = \dfrac{{{{\sin }^2}A}}{{{k^2}}},{b^2} = \dfrac{{{{\sin }^2}B}}{{{k^2}}},{c^2} = \dfrac{{{{\sin }^2}C}}{{{k^2}}}\]
Substitute the values in equation (2)
\[ \Rightarrow \sin A = \dfrac{{\dfrac{{{{\sin }^2}A}}{{{k^2}}}}}{{\dfrac{{{{\sin }^2}B}}{{{k^2}}} - \dfrac{{{{\sin }^2}C}}{{{k^2}}}}}\]
Take LCM in the RHS of the equation.
\[ \Rightarrow \sin A = \dfrac{{\dfrac{{{{\sin }^2}A}}{{{k^2}}}}}{{\dfrac{{{{\sin }^2}B - {{\sin }^2}C}}{{{k^2}}}}}\]
Write the fraction in simple form
\[ \Rightarrow \sin A = \dfrac{{{{\sin }^2}A}}{{{k^2}}} \times \dfrac{{{k^2}}}{{{{\sin }^2}B - {{\sin }^2}C}}\]
Cancel the same terms from numerator and denominator.
\[ \Rightarrow \sin A = \dfrac{{{{\sin }^2}A}}{{{{\sin }^2}B - {{\sin }^2}C}}\]
Use the property \[\sin (x - y)\sin (x + y) = {\sin ^2}x - {\sin ^2}y\]
\[ \Rightarrow \sin A = \dfrac{{{{\sin }^2}A}}{{\sin (B + C).\sin (B - C)}}\] … (3)
Since we know in triangle ABC, the sum of all angles is equal to \[{180^ \circ }\].
Therefore, \[A + B + C = {180^ \circ }\]
Shift angle A to RHS
\[ \Rightarrow B + C = {180^ \circ } - A\]
Apply sin function on both sides of the equation.
\[ \Rightarrow \sin (B + C) = \sin ({180^ \circ } - A)\]
Since, we know \[\sin (\pi - \theta ) = \sin \theta \].
\[ \Rightarrow \sin (B + C) = \sin A\]
Substitute the value of \[\sin (B + C) = \sin A\] in equation (3)
\[ \Rightarrow \sin A = \dfrac{{{{\sin }^2}A}}{{\sin A.\sin (B - C)}}\]
Cross multiply the value of \[\sin A\] from denominator of RHS to LHS
\[ \Rightarrow {\sin ^2}A = \dfrac{{{{\sin }^2}A}}{{\sin (B - C)}}\]
Cancel same terms from numerator of both sides
\[ \Rightarrow \sin (B - C) = 1\]
Substitute the value of \[1 = \sin {90^ \circ }\]in RHS of the equation.
\[ \Rightarrow \sin (B - C) = \sin {90^ \circ }\]
We know sine value on two sides of the equation will be equal if the angles are equal.
\[ \Rightarrow B - C = {90^ \circ }\]
Shift C to RHS
\[ \Rightarrow B = {90^ \circ } + C\]
Substitute the value of C
\[ \Rightarrow B = {90^ \circ } + {23^ \circ }\]
\[ \Rightarrow B = {113^ \circ }\]

So, the correct answer is “Option B”.

Note: Students make mistakes while shifting the values from one side of the equation to another side as they don’t change sign from negative to positive and vice versa. Also, while cross multiplying always brings the value of numerator to denominator and vice versa. Also, whenever writing the value of sine or cos of an angle in a right triangle, take the opposite side to the angle as perpendicular and the largest side as the hypotenuse.