Question
The side BC of a triangle ABC is proceed to D. If ∠ACD = 112° and ∠B = $$\frac{3}{4}$$ ∠A, then the measure of ∠B is:
Answer: Option C
Assume, ∠A = x
∴ ∠B = $$\frac{3}{4}$$ x
∠A + ∠B = 112° (∵ sum of two interior angle is equal to the exterior angle of the third angle)
$$\eqalign{
& {x^ \circ } + \frac{3}{4}{x^ \circ } = {112^ \circ } \cr
& \frac{{7{x^ \circ }}}{4} = {112^ \circ } \cr
& {x^ \circ } = {64^ \circ } \cr
& {\text{Hence,}} \cr
& \angle B = \frac{3}{4} \times {64^ \circ } \cr
& \angle B = {48^ \circ } \cr} $$
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Assume, ∠A = x
∴ ∠B = $$\frac{3}{4}$$ x
∠A + ∠B = 112° (∵ sum of two interior angle is equal to the exterior angle of the third angle)
$$\eqalign{
& {x^ \circ } + \frac{3}{4}{x^ \circ } = {112^ \circ } \cr
& \frac{{7{x^ \circ }}}{4} = {112^ \circ } \cr
& {x^ \circ } = {64^ \circ } \cr
& {\text{Hence,}} \cr
& \angle B = \frac{3}{4} \times {64^ \circ } \cr
& \angle B = {48^ \circ } \cr} $$
Was this answer helpful ?
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