Giiven ΔABC, ∠CAB=75∘ and ∠ACB=45∘. BC is extended to T so that BC=CT. Let M be the midpoint of the segment AT. Find ∠BMC.
The source for this one didn't record the answer, so there is nothing to check what you type against. Work it on paper and mark yourself against the solution below.
Official solution
1. Given ΔABC with ∠CAB=75∘ and ∠ACB=45∘. We need to find ∠BMC where BC is extended to T such that BC=CT and M is the midpoint of AT. 2. Draw the altitude from vertex A to BC and call the intersection point H. This gives us two right triangles ΔABH and ΔAHC. 3. Since ∠CAB=75∘ and ∠ACB=45∘, we can find ∠BAH and ∠HAC: ∠BAH=75∘−45∘=30∘ ∠HAC=45∘ 4. Let BH=x. Then, using trigonometric ratios in ΔABH and ΔAHC: AB=2x(since sin30∘=21) AH=x3(since tan30∘=31) HC=x3(since tan45∘=1) AC=x6(using Pythagoras’ theorem in ΔAHC) 5. Since BC=CT, AM=MT implies M is the midpoint of AT. Therefore, MC is the midsegment of △ABT and MC=x. 6. BM and AC are medians. Let K be the intersection point of BM and AC. K is the centroid of △ABC. 7. The centroid divides each median in the ratio 2:1. Therefore: KCAK=2⟹AK=32x2,KC=3x2 8. Using the Law of Cosines in △KMC: KM2=KC2+CM2−2⋅KC⋅CM⋅cos75∘ We know cos75∘=46−2, so: KM2=(3x2)2+x2−2⋅3x2⋅x⋅46−2 Simplifying the expression: KM2=32x2+x2−23x2(12−6) KM2=32x2+x2−23x2(23−6) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) KM2=32x2+x2−3x2(3−2) \[ KM^2 =
Source: NuminaMath-1.5,
licensed Apache-2.0.
Statement and solution reproduced as published; topic, difficulty and ordering added
by this site.